Class 12- Applied Mathematics-NCERT Handbook Solutions-Unit-3(b)-Integration and its Applications

(i) \( \int (x^2+1)(x-2)dx \)

Expand the integrand:

\[ \int (x^3 - 2x^2 + x - 2) dx \] \[ = \frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} - 2x + C \]

---

(ii) \( \int (x+\frac{1}{x})^2 dx \)

Expand using \( (a+b)^2 \):

\[ \int (x^2 + 2 + \frac{1}{x^2}) dx = \int (x^2 + 2 + x^{-2}) dx \] \[ = \frac{x^3}{3} + 2x + \frac{x^{-1}}{-1} + C \] \[ = \frac{x^3}{3} + 2x - \frac{1}{x} + C \]

---

(iii) \( \int \frac{x^3+x^2+x+1}{x+1} dx \)

Factor the numerator by grouping:

\[ x^3+x^2+x+1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1) \]

Simplify the integral:

\[ \int \frac{(x^2+1)(x+1)}{x+1} dx = \int (x^2+1) dx \] \[ = \frac{x^3}{3} + x + C \]

---

(iv) \( \int \sqrt{3x+5}dx \)

Using the linear substitution rule \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):

\[ = \frac{(3x+5)^{3/2}}{3 \cdot \frac{3}{2}} + C \] \[ = \frac{2}{9}(3x+5)^{3/2} + C \]

---

(v) \( \int (x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2}) dx \)

Use \( (a+b)(a-b) = a^2 - b^2 \):

\[ \int (x^4 - \frac{1}{x^4}) dx = \int (x^4 - x^{-4}) dx \] \[ = \frac{x^5}{5} - \frac{x^{-3}}{-3} + C \] \[ = \frac{x^5}{5} + \frac{1}{3x^3} + C \]

---

(vi) \( \int \frac{1}{\sqrt{x+4}-\sqrt{x-3}} dx \)

Rationalize the denominator:

\[ \int \frac{\sqrt{x+4}+\sqrt{x-3}}{(\sqrt{x+4}-\sqrt{x-3})(\sqrt{x+4}+\sqrt{x-3})} dx \] \[ = \int \frac{\sqrt{x+4}+\sqrt{x-3}}{(x+4)-(x-3)} dx = \int \frac{\sqrt{x+4}+\sqrt{x-3}}{7} dx \] \[ = \frac{1}{7} \left[ \frac{(x+4)^{3/2}}{3/2} + \frac{(x-3)^{3/2}}{3/2} \right] + C \] \[ = \frac{2}{21} \left[ (x+4)^{3/2} + (x-3)^{3/2} \right] + C \]

(i) \( \int \frac{x+e^{2x}}{x^2+e^{2x}} dx \)

Let \( t = x^2 + e^{2x} \). Then \( dt = (2x + 2e^{2x})dx = 2(x+e^{2x})dx \).

So, \( (x+e^{2x})dx = \frac{dt}{2} \).

\[ \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \log|t| + C \] \[ = \frac{1}{2} \log|x^2+e^{2x}| + C \]

---

(ii) \( \int \frac{dx}{\sqrt{x}+x} \)

Rewrite as \( \int \frac{dx}{\sqrt{x}(1+\sqrt{x})} \).

Let \( t = 1 + \sqrt{x} \). Then \( dt = \frac{1}{2\sqrt{x}} dx \), so \( \frac{dx}{\sqrt{x}} = 2dt \).

\[ \int \frac{2dt}{t} = 2 \log|t| + C \] \[ = 2 \log|1+\sqrt{x}| + C \]

---

(iii) \( \int \frac{e^x(1+x)}{(1+xe^x)^2} dx \)

Let \( t = xe^x \). Then \( dt = (e^x \cdot 1 + x \cdot e^x)dx = e^x(1+x)dx \).

\[ \int \frac{dt}{(1+t)^2} = \int (1+t)^{-2} dt \] \[ = \frac{(1+t)^{-1}}{-1} + C = -\frac{1}{1+xe^x} + C \]

---

(iv) \( \int \frac{2x}{\sqrt[3]{x^2+1}} dx \)

Let \( t = x^2 + 1 \). Then \( dt = 2x dx \).

\[ \int \frac{dt}{t^{1/3}} = \int t^{-1/3} dt \] \[ = \frac{t^{2/3}}{2/3} + C = \frac{3}{2}(x^2+1)^{2/3} + C \]

---

(v) \( \int \frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}} dx \)

Let \( t = e^{2x} - e^{-2x} \). Then \( dt = (2e^{2x} + 2e^{-2x})dx = 2(e^{2x}+e^{-2x})dx \).

\[ \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \log|e^{2x}-e^{-2x}| + C \]

---

(vi) \( \int \frac{3e^x-5e^{-x}}{4e^x+5e^{-x}} dx \)

Let \( N = 3e^x-5e^{-x} \) and \( D = 4e^x+5e^{-x} \).

Express \( N = A(D) + B(D') \), where \( D' = 4e^x-5e^{-x} \).

Comparing coefficients:

\[ 4A + 4B = 3 \] \[ 5A - 5B = -5 \Rightarrow A - B = -1 \]

Solving these gives \( A = -1/8 \) and \( B = 7/8 \).

\[ \text{Integral} = A x + B \log|D| + C \] \[ = -\frac{x}{8} + \frac{7}{8} \log|4e^x+5e^{-x}| + C \]

---

(vii) \( \int \frac{2x-3}{x^2-3x-18} dx \)

Let \( t = x^2-3x-18 \). \( dt = (2x-3)dx \).

\[ \int \frac{dt}{t} = \log|x^2-3x-18| + C \]

---

(viii) \( \int \frac{1}{x(1+\log x)^2} dx \)

Let \( t = 1 + \log x \). \( dt = \frac{1}{x} dx \).

\[ \int \frac{dt}{t^2} = -\frac{1}{t} + C = -\frac{1}{1+\log x} + C \]

---

(ix) \( \int \frac{a^{x-1}\cdot\log a + x^{a-1}}{a^x+x^a} dx \)

Let \( t = a^x + x^a \). Then \( dt = (a^x \log a + ax^{a-1})dx = a(a^{x-1}\log a + x^{a-1})dx \).

So, Numerator \( dx = \frac{dt}{a} \).

\[ \int \frac{1}{t} \frac{dt}{a} = \frac{1}{a} \log|a^x+x^a| + C \]

(i) \( \int \frac{1-2x}{\sqrt{1+x^2}} dx \)

Split the integral:

\[ I = \int \frac{1}{\sqrt{1+x^2}} dx - \int \frac{2x}{\sqrt{1+x^2}} dx \]

First part is a standard form: \( \log|x+\sqrt{1+x^2}| \).

Second part: Let \( t = 1+x^2, dt = 2x dx \). \( \int t^{-1/2} dt = 2t^{1/2} = 2\sqrt{1+x^2} \).

\[ I = \log|x+\sqrt{1+x^2}| - 2\sqrt{1+x^2} + C \]

---

(ii) \( \int \frac{1}{\sqrt{3x^2+2x-1}} dx \)

Factor out \( \sqrt{3} \):

\[ \frac{1}{\sqrt{3}} \int \frac{dx}{\sqrt{x^2 + \frac{2}{3}x - \frac{1}{3}}} \]

Complete the square in the denominator:

\[ x^2 + \frac{2}{3}x - \frac{1}{3} = (x + \frac{1}{3})^2 - \frac{1}{9} - \frac{3}{9} = (x + \frac{1}{3})^2 - \left(\frac{2}{3}\right)^2 \]

Using the formula \( \int \frac{dx}{\sqrt{x^2-a^2}} = \log|x+\sqrt{x^2-a^2}| \):

\[ = \frac{1}{\sqrt{3}} \log \left| (x+\frac{1}{3}) + \sqrt{x^2+\frac{2}{3}x-\frac{1}{3}} \right| + C \]

Answer: \( R(x) = 40x - \frac{10}{3}x^3 + 90 \)

Total Revenue \( R(x) \) is the integral of Marginal Revenue.

\[ R(x) = \int (40 - 10x^2) dx = 40x - \frac{10x^3}{3} + C \]

Given \( R = 120 \) when \( x = 3 \):

\[ 120 = 40(3) - \frac{10(3)^3}{3} + C \] \[ 120 = 120 - 90 + C \] \[ 120 = 30 + C \Rightarrow C = 90 \]

Thus, \( R(x) = 40x - \frac{10}{3}x^3 + 90 \).

Answer: \( C(x) = \sqrt{2500+x^2} + 950 \); \( AC(x) = \frac{\sqrt{2500+x^2} + 950}{x} \)

Total Cost Function \( C(x) \):

\[ C(x) = \int \frac{x}{\sqrt{2500+x^2}} dx \]

Let \( t = 2500 + x^2, dt = 2x dx \). \( \int \frac{dt/2}{\sqrt{t}} = \sqrt{t} \).

\[ C(x) = \sqrt{2500+x^2} + K \]

Fixed cost is the cost when output \( x=0 \). Given \( FC = 1000 \).

\[ C(0) = \sqrt{2500} + K = 50 + K \] \[ 1000 = 50 + K \Rightarrow K = 950 \] \[ C(x) = \sqrt{2500+x^2} + 950 \]

Average Cost Function \( AC(x) \):

\[ AC(x) = \frac{C(x)}{x} = \frac{\sqrt{2500+x^2} + 950}{x} \]

Answer: \( C(x) = \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + 7792.53 \)

Cost Function:

\[ C(x) = \int x\sqrt{x+1} dx \]

Let \( u = x+1 \Rightarrow x = u-1, dx = du \).

\[ \int (u-1)u^{1/2} du = \int (u^{3/2} - u^{1/2}) du \] \[ = \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + K \] \[ C(x) = \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + K \]

Given \( C(3) = 7800 \):

\[ 7800 = \frac{2}{5}(4)^{5/2} - \frac{2}{3}(4)^{3/2} + K \] \[ 7800 = \frac{2}{5}(32) - \frac{2}{3}(8) + K \] \[ 7800 = 12.8 - 5.33 + K \Rightarrow 7800 = 7.47 + K \] \[ K \approx 7792.53 \] \[ C(x) = \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + 7792.53 \]

(i) \( \int \frac{x+1}{(x+2)(x+4)} dx \)

Using partial fractions:

\[ \frac{x+1}{(x+2)(x+4)} = \frac{A}{x+2} + \frac{B}{x+4} \] \[ x+1 = A(x+4) + B(x+2) \]

At \( x = -2 \): \( -1 = 2A \Rightarrow A = -\frac{1}{2} \).

At \( x = -4 \): \( -3 = -2B \Rightarrow B = \frac{3}{2} \).

\[ \text{Integral} = -\frac{1}{2}\log|x+2| + \frac{3}{2}\log|x+4| + C \]

---

(ii) \( \int \frac{x}{(x^2+1)(x^2+2)} dx \)

Let \( x^2 = t \), then \( 2x dx = dt \Rightarrow x dx = \frac{dt}{2} \).

\[ \frac{1}{2} \int \frac{dt}{(t+1)(t+2)} = \frac{1}{2} \int \left( \frac{1}{t+1} - \frac{1}{t+2} \right) dt \] \[ = \frac{1}{2} [\log|t+1| - \log|t+2|] + C \] \[ = \frac{1}{2} \log\left| \frac{x^2+1}{x^2+2} \right| + C \]

---

(iii) \( \int \frac{1}{e^{2x}-1} dx \)

Let \( e^x = t \), \( e^x dx = dt \Rightarrow dx = \frac{dt}{t} \).

\[ \int \frac{1}{t^2-1} \cdot \frac{dt}{t} = \int \frac{1}{t(t-1)(t+1)} dt \]

Partial fractions: \( \frac{1}{t(t^2-1)} = -\frac{1}{t} + \frac{1}{2(t-1)} + \frac{1}{2(t+1)} \).

\[ \text{Integral} = -\log|t| + \frac{1}{2}\log|t-1| + \frac{1}{2}\log|t+1| + C \] \[ = -x + \frac{1}{2}\log|e^{2x}-1| + C \]

---

(iv) \( \int \frac{1}{x((\log x)^2 - 3\log x + 2)} dx \)

Let \( \log x = t \), \( \frac{1}{x} dx = dt \).

\[ \int \frac{dt}{t^2-3t+2} = \int \frac{dt}{(t-1)(t-2)} \] \[ = \int \left( \frac{1}{t-2} - \frac{1}{t-1} \right) dt = \log|t-2| - \log|t-1| + C \] \[ = \log\left| \frac{\log x - 2}{\log x - 1} \right| + C \]

---

(v) \( \int \frac{3x-2}{(x-2)^2(x+2)} dx \)

Form: \( \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+2} \).

Solving gives \( A = \frac{11}{32} \), \( B = 1 \), \( C = -\frac{11}{32} \). (Wait, let's recheck quickly: \( x=2 \Rightarrow 4 = 4B \Rightarrow B=1 \). \( x=-2 \Rightarrow -8 = 16C \Rightarrow C=-1/2 \). \( A+C=0 \Rightarrow A=1/2 \). Correct values: \( A=1/2, B=1, C=-1/2 \)).

\[ \text{Integral} = \frac{1}{2}\log|x-2| - \frac{1}{x-2} - \frac{1}{2}\log|x+2| + C \]

---

(vi) \( \int \frac{1}{e^{2x} + e^x} dx \)

Let \( e^x = t \), \( dx = \frac{dt}{t} \).

\[ \int \frac{1}{t^2+t} \frac{dt}{t} \text{ (Incorrect sub step, correction: } \int \frac{1}{t(t+1)} \frac{dt}{t} \text{? No)} \]

Correction: \( \int \frac{1}{e^x(e^x+1)} dx \). Let \( e^x=t \), \( dx=dt/t \). Integral is \( \int \frac{1}{t^2(t+1)} dt \).

Partial fractions on \( \frac{1}{t^2(t+1)} = -\frac{1}{t} + \frac{1}{t^2} + \frac{1}{t+1} \).

\[ = -\log|t| - \frac{1}{t} + \log|t+1| + C \] \[ = \log(e^x+1) - x - e^{-x} + C \]

Alternative approach: \( \int \frac{e^{-2x}}{1+e^{-x}} dx \). Or simple partial fractions.

---

(vii) \( \int \frac{5x+4}{(x^2-1)(x+2)} dx \)

Factor denominator: \( (x-1)(x+1)(x+2) \).

\[ \frac{5x+4}{(x-1)(x+1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x+2} \]

\( A = 3/2 \), \( B = 1/2 \), \( C = -2 \).

\[ = \frac{3}{2}\log|x-1| + \frac{1}{2}\log|x+1| - 2\log|x+2| + C \]

---

(viii) \( \int \frac{x}{(x-1)^2(x+2)} dx \)

\( A/(x-1) + B/(x-1)^2 + C/(x+2) \).

\( B = 1/3 \), \( C = -2/9 \), \( A = 2/9 \).

\[ = \frac{2}{9}\log|x-1| - \frac{1}{3(x-1)} - \frac{2}{9}\log|x+2| + C \]

---

(ix) \( \int \frac{1}{x(x^4-1)} dx \)

Multiply numerator and denominator by \( x^3 \): \( \int \frac{x^3}{x^4(x^4-1)} dx \).

Let \( x^4 = t \), \( 4x^3 dx = dt \).

\[ \frac{1}{4} \int \frac{dt}{t(t-1)} = \frac{1}{4} (\log|t-1| - \log|t|) + C \] \[ = \frac{1}{4} \log\left| \frac{x^4-1}{x^4} \right| + C \]

---

(x) \( \int \frac{1}{x(x^n+1)} dx \)

Similar to (ix), multiply by \( x^{n-1} \). Let \( x^n = t \).

\[ = \frac{1}{n} \log\left| \frac{x^n}{x^n+1} \right| + C \]

---

(xi) \( \int \frac{1-x}{x(1-2x)} dx \)

Partial fractions: \( \frac{1-x}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x} \).

\( 1-x = A(1-2x) + Bx \).

\( x=0 \Rightarrow A=1 \). \( x=1/2 \Rightarrow 1/2 = B/2 \Rightarrow B=1 \).

\[ \int \left( \frac{1}{x} + \frac{1}{1-2x} \right) dx = \log|x| - \frac{1}{2}\log|1-2x| + C \]

Proof:

Revenue \( R(x) = \int MR(x) dx \).

\[ MR(x) = \frac{5x^2+30x+51}{(x+3)^2} = \frac{5(x^2+6x+9) + 6}{(x+3)^2} \] \[ = \frac{5(x+3)^2 + 6}{(x+3)^2} = 5 + \frac{6}{(x+3)^2} \]

Integrate w.r.t \( x \):

\[ R(x) = \int \left( 5 + 6(x+3)^{-2} \right) dx = 5x + 6 \frac{(x+3)^{-1}}{-1} + C \] \[ R(x) = 5x - \frac{6}{x+3} + C \]

When output \( x = 0 \), Revenue \( R = 0 \).

\[ 0 = 0 - \frac{6}{3} + C \Rightarrow -2 + C = 0 \Rightarrow C = 2 \]

So, \( R(x) = 5x - \frac{6}{x+3} + 2 \).

To show it equals \( \frac{2x}{x+3} + 5x \), we simplify the constant part:

\[ 2 - \frac{6}{x+3} = \frac{2(x+3)-6}{x+3} = \frac{2x+6-6}{x+3} = \frac{2x}{x+3} \]

Thus, \( R(x) = 5x + \frac{2x}{x+3} \). Hence Proved.

Answer:

1. Total Revenue Function \( R(x) \):

\[ R(x) = \int MR(x) dx = \int \left( ab(x+b)^{-2} - c \right) dx \] \[ = ab \frac{(x+b)^{-1}}{-1} - cx + K \] \[ = -\frac{ab}{x+b} - cx + K \]

At \( x=0, R=0 \):

\[ 0 = -\frac{ab}{b} - 0 + K \Rightarrow K = a \]

Substitute \( K \):

\[ R(x) = a - \frac{ab}{x+b} - cx = \frac{a(x+b) - ab}{x+b} - cx \] \[ R(x) = \frac{ax}{x+b} - cx \]

2. Demand Function \( p(x) \):

Demand function \( p = \frac{R(x)}{x} \).

\[ p = \frac{1}{x} \left( \frac{ax}{x+b} - cx \right) = \frac{a}{x+b} - c \]

(i) \( \int x e^{2x+3} dx \)

Using integration by parts: \( \int u dv = uv - \int v du \).

Let \( u = x \Rightarrow du = dx \).

Let \( dv = e^{2x+3} dx \Rightarrow v = \frac{1}{2}e^{2x+3} \).

\[ I = \frac{x}{2}e^{2x+3} - \int \frac{1}{2}e^{2x+3} dx \] \[ I = \frac{x}{2}e^{2x+3} - \frac{1}{4}e^{2x+3} + C = \frac{e^{2x+3}}{4}(2x-1) + C \]

---

(ii) \( \int x \log(x^2 + 1) dx \)

Substitution: Let \( x^2 + 1 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2} \).

\[ I = \frac{1}{2} \int \log t \, dt \]

Standard integral of \( \log t \) is \( t \log t - t \):

\[ I = \frac{1}{2} [t \log t - t] + C \] \[ I = \frac{1}{2} [(x^2+1)\log(x^2+1) - (x^2+1)] + C \]

---

(iii) \( \int x^2 e^x dx \)

Integration by parts twice.

1st: \( u = x^2, dv = e^x dx \Rightarrow I = x^2 e^x - \int 2x e^x dx \).

2nd: For \( \int x e^x dx \), let \( u = x, dv = e^x dx \Rightarrow x e^x - e^x \).

\[ I = x^2 e^x - 2(x e^x - e^x) + C \] \[ I = e^x(x^2 - 2x + 2) + C \]

---

(iv) \( \int x \log x dx \)

Let \( u = \log x \) (ILATE rule), \( dv = x dx \).

\( du = \frac{1}{x} dx, v = \frac{x^2}{2} \).

\[ I = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx \] \[ I = \frac{x^2}{2} \log x - \frac{1}{2} \int x dx \] \[ I = \frac{x^2}{2} \log x - \frac{x^2}{4} + C \]

---

(v) \( \int x \log 2x dx \)

Let \( u = \log 2x, dv = x dx \).

\( du = \frac{1}{2x} \cdot 2 dx = \frac{1}{x} dx, v = \frac{x^2}{2} \).

\[ I = \frac{x^2}{2} \log 2x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx \] \[ I = \frac{x^2}{2} \log 2x - \frac{x^2}{4} + C \]

---

(vi) \( \int x^2 \log x dx \)

Let \( u = \log x, dv = x^2 dx \).

\( du = \frac{1}{x} dx, v = \frac{x^3}{3} \).

\[ I = \frac{x^3}{3} \log x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx \] \[ I = \frac{x^3}{3} \log x - \frac{1}{3} \int x^2 dx \] \[ I = \frac{x^3}{3} \log x - \frac{x^3}{9} + C \]

---

(vii) \( \int (x^2 + 1) \log x dx \)

Split into \( \int x^2 \log x dx + \int \log x dx \).

From (vi): \( \int x^2 \log x dx = \frac{x^3}{3}\log x - \frac{x^3}{9} \).

Standard integral: \( \int \log x dx = x \log x - x \).

\[ I = \frac{x^3}{3}\log x - \frac{x^3}{9} + x \log x - x + C \] \[ I = \left(\frac{x^3}{3} + x\right)\log x - \left(\frac{x^3}{9} + x\right) + C \]

---

(viii) \( \int x (\log x)^2 dx \)

Let \( u = (\log x)^2, dv = x dx \).

\( du = 2 \log x \cdot \frac{1}{x} dx, v = \frac{x^2}{2} \).

\[ I = \frac{x^2}{2}(\log x)^2 - \int \frac{x^2}{2} \cdot \frac{2 \log x}{x} dx \] \[ I = \frac{x^2}{2}(\log x)^2 - \int x \log x dx \]

Using result from (iv):

\[ I = \frac{x^2}{2}(\log x)^2 - \left( \frac{x^2}{2} \log x - \frac{x^2}{4} \right) + C \]

(i) \( \int e^x [x^{-2} - 2x^{-3}] dx \)

Using the property \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \).

Let \( f(x) = x^{-2} \). Then \( f'(x) = -2x^{-3} \).

This matches the integrand exactly.

\[ I = e^x x^{-2} + C = \frac{e^x}{x^2} + C \]

---

(ii) \( \int e^{3x} [\log 3x + \frac{1}{3x}] dx \)

Let \( 3x = t \Rightarrow 3dx = dt \). Integral becomes \( \frac{1}{3} \int e^t [\log t + \frac{1}{t}] dt \).

Here \( f(t) = \log t \) and \( f'(t) = \frac{1}{t} \).

\[ I = \frac{1}{3} e^t \log t + C = \frac{1}{3} e^{3x} \log 3x + C \]

---

(iii) \( \int e^{2x} [\frac{2x-1}{4x^2}] dx \)

Simplify term inside bracket: \( \frac{2x}{4x^2} - \frac{1}{4x^2} = \frac{1}{2x} - \frac{1}{4x^2} \).

Let \( 2x = t \). Then \( \frac{1}{2x} = \frac{1}{t} \) and \( \frac{1}{4x^2} = \frac{1}{t^2} \). Also \( dx = \frac{dt}{2} \).

\[ I = \int e^t \left[ \frac{1}{t} - \frac{1}{t^2} \right] \frac{dt}{2} \]

Using \( f(t) = \frac{1}{t}, f'(t) = -\frac{1}{t^2} \):

\[ I = \frac{1}{2} e^t \left(\frac{1}{t}\right) + C = \frac{e^{2x}}{2(2x)} + C = \frac{e^{2x}}{4x} + C \]

---

(iv) \( \int [\log(\log x) + \frac{1}{(\log x)^2}] dx \)

Let \( \log x = t \Rightarrow x = e^t \Rightarrow dx = e^t dt \).

\[ I = \int \left[ \log t + \frac{1}{t^2} \right] e^t dt \]

Add and subtract \( \frac{1}{t} \):

\[ I = \int e^t \left[ \left(\log t + \frac{1}{t}\right) - \left(\frac{1}{t} - \frac{1}{t^2}\right) \right] dt \]

First part: \( f(t) = \log t, f'(t) = 1/t \Rightarrow e^t \log t \).

Second part: \( g(t) = 1/t, g'(t) = -1/t^2 \Rightarrow e^t (1/t) \).

\[ I = e^t \log t - \frac{e^t}{t} + C \] \[ I = x \log(\log x) - \frac{x}{\log x} + C \]