Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.4

Solution:

Let \( x^3 = t \). Then \( 3x^2 \, dx = dt \).

Substitute into the integral:

\[ \begin{aligned} \int \frac{3x^2}{(x^3)^2+1} \, dx &= \int \frac{dt}{t^2+1} \\ &= \tan^{-1} t + C \\ &= \tan^{-1}(x^3) + C \end{aligned} \]

Solution:

Rewrite as \( \frac{1}{\sqrt{1+(2x)^2}} \).

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

\[ \begin{aligned} \int \frac{1}{\sqrt{1+t^2}} \cdot \frac{dt}{2} &= \frac{1}{2} \log |t + \sqrt{1+t^2}| + C \\ &= \frac{1}{2} \log |2x + \sqrt{1+4x^2}| + C \end{aligned} \]

Solution:

Let \( 2-x = t \Rightarrow -dx = dt \Rightarrow dx = -dt \).

\[ \begin{aligned} \int \frac{-dt}{\sqrt{t^2+1}} &= -\log |t + \sqrt{t^2+1}| + C \\ &= -\log |2-x + \sqrt{(2-x)^2+1}| + C \\ &= \log \left| \frac{1}{2-x+\sqrt{x^2-4x+5}} \right| + C \end{aligned} \]

Solution:

Rewrite as \( \frac{1}{\sqrt{3^2-(5x)^2}} \).

Let \( 5x = t \Rightarrow 5 \, dx = dt \Rightarrow dx = \frac{dt}{5} \).

\[ \begin{aligned} \int \frac{1}{\sqrt{3^2-t^2}} \cdot \frac{dt}{5} &= \frac{1}{5} \sin^{-1} \left( \frac{t}{3} \right) + C \\ &= \frac{1}{5} \sin^{-1} \left( \frac{5x}{3} \right) + C \end{aligned} \]

Solution:

Rewrite as \( \frac{3x}{1+(\sqrt{2}x^2)^2} \).

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

\[ \begin{aligned} \int \frac{3}{1+2t^2} \cdot \frac{dt}{2} &= \frac{3}{2} \int \frac{dt}{1+(\sqrt{2}t)^2} \\ \text{Let } \sqrt{2}t = u \Rightarrow \sqrt{2}dt = du \\ &= \frac{3}{2\sqrt{2}} \int \frac{du}{1+u^2} \\ &= \frac{3}{2\sqrt{2}} \tan^{-1} u + C \\ &= \frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2}x^2) + C \end{aligned} \]

Solution:

Let \( x^3 = t \Rightarrow 3x^2 \, dx = dt \Rightarrow x^2 \, dx = \frac{dt}{3} \).

\[ \begin{aligned} \int \frac{1}{1-t^2} \cdot \frac{dt}{3} &= \frac{1}{3} \cdot \frac{1}{2(1)} \log \left| \frac{1+t}{1-t} \right| + C \\ &= \frac{1}{6} \log \left| \frac{1+x^3}{1-x^3} \right| + C \end{aligned} \]

Solution:

Split the integral:

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

For the first part, let \( x^2-1 = t \Rightarrow 2x \, dx = dt \).

\[ \begin{aligned} I_1 &= \frac{1}{2} \int t^{-1/2} \, dt = \sqrt{t} = \sqrt{x^2-1} \\ I_2 &= \int \frac{dx}{\sqrt{x^2-1}} = \log |x+\sqrt{x^2-1}| \\ \text{Result: } & \sqrt{x^2-1} - \log |x+\sqrt{x^2-1}| + C \end{aligned} \]

Solution:

Let \( x^3 = t \Rightarrow 3x^2 \, dx = dt \).

\[ \begin{aligned} \int \frac{1}{\sqrt{t^2+(a^3)^2}} \cdot \frac{dt}{3} &= \frac{1}{3} \log |t + \sqrt{t^2+a^6}| + C \\ &= \frac{1}{3} \log |x^3 + \sqrt{x^6+a^6}| + C \end{aligned} \]

Solution:

Let \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \).

\[ \begin{aligned} \int \frac{dt}{\sqrt{t^2+2^2}} &= \log |t + \sqrt{t^2+4}| + C \\ &= \log |\tan x + \sqrt{\tan^2 x + 4}| + C \end{aligned} \]

Solution:

Complete the square: \( x^2+2x+2 = (x+1)^2 + 1 \).

\[ \begin{aligned} \int \frac{dx}{\sqrt{(x+1)^2+1^2}} &= \log |(x+1) + \sqrt{(x+1)^2+1}| + C \\ &= \log |x+1 + \sqrt{x^2+2x+2}| + C \end{aligned} \]

Solution:

Factor out 9: \( \frac{1}{9(x^2+\frac{2}{3}x+\frac{5}{9})} \).

Complete square: \( x^2+\frac{2}{3}x+\frac{5}{9} = (x+\frac{1}{3})^2 + \frac{4}{9} = (x+\frac{1}{3})^2 + (\frac{2}{3})^2 \).

\[ \begin{aligned} \frac{1}{9} \int \frac{dx}{(x+1/3)^2 + (2/3)^2} &= \frac{1}{9} \cdot \frac{1}{2/3} \tan^{-1} \left( \frac{x+1/3}{2/3} \right) + C \\ &= \frac{1}{6} \tan^{-1} \left( \frac{3x+1}{2} \right) + C \end{aligned} \]

Solution:

Rewrite expression inside root: \( -(x^2+6x-7) = -((x+3)^2-9-7) = -( (x+3)^2 - 16 ) = 16 - (x+3)^2 = 4^2 - (x+3)^2 \).

\[ \begin{aligned} \int \frac{dx}{\sqrt{4^2-(x+3)^2}} &= \sin^{-1} \left( \frac{x+3}{4} \right) + C \end{aligned} \]

Solution:

\( (x-1)(x-2) = x^2-3x+2 \).

Complete square: \( (x-\frac{3}{2})^2 - \frac{9}{4} + 2 = (x-\frac{3}{2})^2 - \frac{1}{4} \).

\[ \begin{aligned} \int \frac{dx}{\sqrt{(x-3/2)^2 - (1/2)^2}} &= \log |(x-3/2) + \sqrt{x^2-3x+2}| + C \\ &= \log |x - \frac{3}{2} + \sqrt{x^2-3x+2}| + C \end{aligned} \]

Solution:

\( 8+3x-x^2 = -(x^2-3x-8) = -((x-3/2)^2 - 9/4 - 32/4) = -((x-3/2)^2 - 41/4) = (\frac{\sqrt{41}}{2})^2 - (x-\frac{3}{2})^2 \).

\[ \begin{aligned} \int \frac{dx}{\sqrt{(\sqrt{41}/2)^2 - (x-3/2)^2}} &= \sin^{-1} \left( \frac{x-3/2}{\sqrt{41}/2} \right) + C \\ &= \sin^{-1} \left( \frac{2x-3}{\sqrt{41}} \right) + C \end{aligned} \]

Solution:

Quadratic form: \( x^2 - (a+b)x + ab = (x - \frac{a+b}{2})^2 - (\frac{a-b}{2})^2 \).

\[ \begin{aligned} \int \frac{dx}{\sqrt{(x - \frac{a+b}{2})^2 - (\frac{a-b}{2})^2}} &= \log |x - \frac{a+b}{2} + \sqrt{(x-a)(x-b)}| + C \end{aligned} \]

Solution:

Let \( 2x^2+x-3 = t \Rightarrow (4x+1)dx = dt \).

\[ \begin{aligned} \int \frac{dt}{\sqrt{t}} &= 2\sqrt{t} + C \\ &= 2\sqrt{2x^2+x-3} + C \end{aligned} \]

Solution:

Split into \( \frac{x}{\sqrt{x^2-1}} + \frac{2}{\sqrt{x^2-1}} \).

\[ \begin{aligned} I_1 &= \int \frac{x}{\sqrt{x^2-1}} dx = \sqrt{x^2-1} \\ I_2 &= 2 \int \frac{dx}{\sqrt{x^2-1}} = 2 \log |x+\sqrt{x^2-1}| \\ \text{Result: } & \sqrt{x^2-1} + 2 \log |x+\sqrt{x^2-1}| + C \end{aligned} \]

Solution:

Express numerator as \( A \frac{d}{dx}(\text{denominator}) + B \).

\( 5x-2 = A(6x+2) + B \Rightarrow 6A=5, 2A+B=-2 \Rightarrow A=\frac{5}{6}, B=-\frac{11}{3} \).

\[ I = \frac{5}{6} \int \frac{6x+2}{1+2x+3x^2} dx - \frac{11}{3} \int \frac{dx}{3x^2+2x+1} \]

First part: \( \frac{5}{6} \log|1+2x+3x^2| \).

Second part: \( \frac{11}{3} \int \frac{dx}{3(x+1/3)^2 + 2/3} = \frac{11}{9} \int \frac{dx}{(x+1/3)^2 + (\sqrt{2}/3)^2} \).

\( = \frac{11}{9} \cdot \frac{1}{\sqrt{2}/3} \tan^{-1} \frac{x+1/3}{\sqrt{2}/3} = \frac{11}{3\sqrt{2}} \tan^{-1} \frac{3x+1}{\sqrt{2}} \).

Result: \( \frac{5}{6} \log|1+2x+3x^2| - \frac{11}{3\sqrt{2}} \tan^{-1} \left( \frac{3x+1}{\sqrt{2}} \right) + C \).

Solution:

Denominator is \( \sqrt{x^2-9x+20} \).

\( 6x+7 = A(2x-9) + B \Rightarrow A=3, B=34 \).

\[ I = 3 \int \frac{2x-9}{\sqrt{x^2-9x+20}} dx + 34 \int \frac{dx}{\sqrt{(x-9/2)^2 - 1/4}} \]

Result: \( 3 \cdot 2\sqrt{x^2-9x+20} + 34 \log |x-9/2 + \sqrt{x^2-9x+20}| + C \).

\( = 6\sqrt{x^2-9x+20} + 34 \log |x-4.5 + \sqrt{x^2-9x+20}| + C \).

Solution:

\( x+2 = A(4-2x) + B \Rightarrow -2A=1 (A=-1/2), 4A+B=2 (B=4) \).

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

Result: \( -\sqrt{4x-x^2} + 4 \sin^{-1} \left( \frac{x-2}{2} \right) + C \).

Solution:

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

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

Result: \( \sqrt{x^2+2x+3} + \log |x+1 + \sqrt{x^2+2x+3}| + C \).

Solution:

\( x+3 = A(2x-2) + B \Rightarrow A=1/2, B=4 \).

\[ I = \frac{1}{2} \log |x^2-2x-5| + 4 \int \frac{dx}{(x-1)^2-(\sqrt{6})^2} \]

Result: \( \frac{1}{2} \log |x^2-2x-5| + \frac{4}{2\sqrt{6}} \log \left| \frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| + C \).

Solution:

\( 5x+3 = A(2x+4) + B \Rightarrow A=5/2, B=-7 \).

\[ I = \frac{5}{2} \cdot 2\sqrt{x^2+4x+10} - 7 \int \frac{dx}{\sqrt{(x+2)^2+6}} \]

Result: \( 5\sqrt{x^2+4x+10} - 7 \log |x+2 + \sqrt{x^2+4x+10}| + C \).

Answer: (B)

Solution:

\( \int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1) + C \).

Answer: (B)

Solution:

\( \int \frac{dx}{\sqrt{4(9x/4-x^2)}} = \frac{1}{2} \int \frac{dx}{\sqrt{-(x^2-9x/4)}} \).

Inside root: \( -((x-9/8)^2 - 81/64) = (9/8)^2 - (x-9/8)^2 \).

Integral: \( \frac{1}{2} \sin^{-1} \left( \frac{x-9/8}{9/8} \right) = \frac{1}{2} \sin^{-1} \left( \frac{8x-9}{9} \right) + C \).