Where Mindful Learning Begins
Where Mindful Learning Begins
These Class 12 Maths NCERT Solutions Chapter 7 Ex 7.4 cover all 25 questions — integrals of particular functions using completing the square and splitting the numerator to match the six standard forms.
Every question in this exercise is really a pattern-matching exercise in disguise. NCERT gives you six standard integral formulae involving expressions like a² − x², x² + a², and their square-root versions — the entire skill here is manipulating the given integrand until it fits one of those six forms. When the denominator or the term under a square root is a quadratic, completing the square turns it into (x + k)² ± a constant, which matches directly. When the numerator is linear and the denominator is a quadratic (or its square root), splitting the numerator into a multiple of the denominator's derivative plus a constant separates the problem into an easy substitution part and a standard-formula part. The last two questions are MCQs that test the same skill under exam time pressure. This exercise is dense, but mastering it makes Ex 7.4-style questions some of the fastest marks available on the board paper.
\dfrac{3x^2}{x^6+1} = \dfrac{3x^2}{(x^3)^2+1}
Put x^3=t, so 3x^2\,dx=dt
\displaystyle\int \dfrac{3x^2}{x^6+1}\,dx = \int \dfrac{dt}{t^2+1} = \tan^{-1}t [∵ \displaystyle\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=1]
\dfrac{1}{\sqrt{1+4x^2}} = \dfrac{1}{\sqrt{1+(2x)^2}}
Put 2x=t, so 2\,dx=dt
\displaystyle\int \dfrac{1}{\sqrt{1+4x^2}}\,dx = \dfrac12\int \dfrac{dt}{\sqrt{t^2+1}} = \dfrac12\log\left|t+\sqrt{t^2+1}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|\text{, here }a=1]
Put 2-x=t, so -dx=dt
\displaystyle\int \dfrac{1}{\sqrt{(2-x)^2+1}}\,dx = -\int \dfrac{dt}{\sqrt{t^2+1}} = -\log\left|t+\sqrt{t^2+1}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|\text{, here }a=1]
\dfrac{1}{\sqrt{9-25x^2}} = \dfrac{1}{\sqrt{3^2-(5x)^2}}
Put 5x=t, so 5\,dx=dt
\displaystyle\int \dfrac{1}{\sqrt{9-25x^2}}\,dx = \dfrac15\int \dfrac{dt}{\sqrt{3^2-t^2}} = \dfrac15\sin^{-1}\left(\dfrac{t}{3}\right) [∵ \displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=3]
Put x^2=t, so 2x\,dx=dt, i.e. 3x\,dx=\dfrac32\,dt. The denominator becomes 1+2t^2
\displaystyle\int \dfrac{3x}{1+2x^4}\,dx = \dfrac32\int \dfrac{dt}{1+2t^2} = \dfrac32\cdot\dfrac12\int \dfrac{dt}{t^2+\frac12} = \dfrac34\cdot\sqrt{2}\,\tan^{-1}(\sqrt2\,t) [∵ \displaystyle\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=\dfrac{1}{\sqrt2}]
\dfrac{x^2}{1-x^6} = \dfrac{x^2}{1-(x^3)^2}
Put x^3=t, so 3x^2\,dx=dt
\displaystyle\int \dfrac{x^2}{1-x^6}\,dx = \dfrac13\int \dfrac{dt}{1-t^2} = \dfrac13\cdot\dfrac12\log\left|\dfrac{1+t}{1-t}\right| [∵ \displaystyle\int \dfrac{dx}{a^2-x^2} = \dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|\text{, here }a=1]
Let I = \displaystyle\int \dfrac{x-1}{\sqrt{x^2-1}}\,dx
\dfrac{x-1}{\sqrt{x^2-1}} = \dfrac{x}{\sqrt{x^2-1}} - \dfrac{1}{\sqrt{x^2-1}}
I = \displaystyle\int \dfrac{x}{\sqrt{x^2-1}}\,dx - \int \dfrac{dx}{\sqrt{x^2-1}}
For the first integral, put x^2-1=u, so 2x\,dx=du: \displaystyle\int \dfrac{x}{\sqrt{x^2-1}}\,dx = \int \dfrac{du}{2\sqrt u} = \sqrt{x^2-1}
The second integral is a standard result: \displaystyle\int \dfrac{dx}{\sqrt{x^2-1}} = \log\left|x+\sqrt{x^2-1}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right|\text{, here }a=1]
\dfrac{x^2}{\sqrt{x^6+a^6}} = \dfrac{x^2}{\sqrt{(x^3)^2+(a^3)^2}}
Put x^3=t, so 3x^2\,dx=dt
\displaystyle\int \dfrac{x^2}{\sqrt{x^6+a^6}}\,dx = \dfrac13\int \dfrac{dt}{\sqrt{t^2+(a^3)^2}} = \dfrac13\log\left|t+\sqrt{t^2+a^6}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2+k^2}} = \log\left|x+\sqrt{x^2+k^2}\right|\text{, here }k=a^3]
Put \tan x=t, so \sec^2x\,dx=dt
\displaystyle\int \dfrac{\sec^2 x}{\sqrt{\tan^2 x+4}}\,dx = \int \dfrac{dt}{\sqrt{t^2+4}} = \log\left|t+\sqrt{t^2+4}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|\text{, here }a=2]
Complete the square: x^2+2x+2 = (x+1)^2+1
Put x+1=t: \displaystyle\int \dfrac{1}{\sqrt{x^2+2x+2}}\,dx = \int \dfrac{dt}{\sqrt{t^2+1}} = \log\left|t+\sqrt{t^2+1}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|\text{, here }a=1]
Let I = \displaystyle\int \dfrac{1}{9x^2+6x+5}\,dx
Since the coefficient of x^2 is not 1, factor it out first: 9x^2+6x+5 = 9\left[x^2+\dfrac23 x\right]+5
Complete the square inside the bracket: 9\left[x^2+\dfrac23 x+\left(\dfrac13\right)^2-\left(\dfrac13\right)^2\right]+5 = 9\left[\left(x+\dfrac13\right)^2-\dfrac19\right]+5
= 9\left(x+\dfrac13\right)^2-1+5 = 9\left(x+\dfrac13\right)^2+4
Put x+\dfrac13=t: I = \displaystyle\int \dfrac{dt}{9t^2+4} = \dfrac19\int \dfrac{dt}{t^2+\frac49} = \dfrac16\tan^{-1}\left(\dfrac{3t}{2}\right) [∵ \displaystyle\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=\dfrac23]
Complete the square: 7-6x-x^2 = 16-(x+3)^2
Put x+3=t: \displaystyle\int \dfrac{1}{\sqrt{7-6x-x^2}}\,dx = \int \dfrac{dt}{\sqrt{16-t^2}} = \sin^{-1}\left(\dfrac{t}{4}\right) [∵ \displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=4]
Expand and complete the square: (x-1)(x-2)=x^2-3x+2 = \left(x-\dfrac32\right)^2-\dfrac14
Put x-\dfrac32=t: \displaystyle\int \dfrac{1}{\sqrt{(x-1)(x-2)}}\,dx = \int \dfrac{dt}{\sqrt{t^2-\frac14}} = \log\left|t+\sqrt{t^2-\frac14}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right|\text{, here }a=\dfrac12]
Complete the square: 8+3x-x^2 = \dfrac{41}{4}-\left(x-\dfrac32\right)^2
Put x-\dfrac32=t: \displaystyle\int \dfrac{1}{\sqrt{8+3x-x^2}}\,dx = \int \dfrac{dt}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-t^2}} = \sin^{-1}\left(\dfrac{2t}{\sqrt{41}}\right) [∵ \displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=\dfrac{\sqrt{41}}{2}]
Expand and complete the square: (x-a)(x-b) = \left(x-\dfrac{a+b}{2}\right)^2-\left(\dfrac{a-b}{2}\right)^2
Put x-\dfrac{a+b}{2}=t: \displaystyle\int \dfrac{1}{\sqrt{(x-a)(x-b)}}\,dx = \int \dfrac{dt}{\sqrt{t^2-\left(\frac{a-b}{2}\right)^2}} = \log\left|t+\sqrt{t^2-\left(\frac{a-b}{2}\right)^2}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2-k^2}} = \log\left|x+\sqrt{x^2-k^2}\right|\text{, here }k=\dfrac{a-b}{2}]
\dfrac{d}{dx}(2x^2+x-3) = 4x+1 — exactly the numerator
Put 2x^2+x-3=t, so (4x+1)\,dx=dt: \displaystyle\int \dfrac{4x+1}{\sqrt{2x^2+x-3}}\,dx = \int \dfrac{dt}{\sqrt{t}} = 2\sqrt{t} [∵ \displaystyle\int \dfrac{f'(x)}{\sqrt{f(x)}}\,dx = 2\sqrt{f(x)}]
Express x+2 = A\cdot\dfrac{d}{dx}(x^2-1)+B = A(2x)+B
Comparing coefficients: A=\dfrac12, B=2
\displaystyle\int \dfrac{x+2}{\sqrt{x^2-1}}\,dx = \dfrac12\int \dfrac{2x}{\sqrt{x^2-1}}\,dx + 2\int \dfrac{dx}{\sqrt{x^2-1}}
= \dfrac12\cdot2\sqrt{x^2-1} + 2\log\left|x+\sqrt{x^2-1}\right| [∵ \displaystyle\int \dfrac{f'}{\sqrt f}\,dx = 2\sqrt f\text{, and }\int \dfrac{dx}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right|\text{, here }a=1]
Express 5x-2 = A\cdot\dfrac{d}{dx}(3x^2+2x+1)+B = A(6x+2)+B
Comparing coefficients: 6A=5\Rightarrow A=\dfrac56, and 2A+B=-2\Rightarrow B=-\dfrac{11}{3}
First part: \displaystyle\dfrac56\int \dfrac{6x+2}{3x^2+2x+1}\,dx = \dfrac56\log(3x^2+2x+1) [∵ \displaystyle\int \dfrac{f'}{f}\,dx = \log|f|]
For the second part, the coefficient of x^2 is not 1, so factor it out first: 3x^2+2x+1 = 3\left[x^2+\dfrac23 x\right]+1
Complete the square inside the bracket: 3\left[x^2+\dfrac23 x+\left(\dfrac13\right)^2-\left(\dfrac13\right)^2\right]+1 = 3\left[\left(x+\dfrac13\right)^2-\dfrac19\right]+1 = 3\left(x+\dfrac13\right)^2-\dfrac13+1 = 3\left(x+\dfrac13\right)^2+\dfrac23
So \displaystyle\int \dfrac{dx}{3x^2+2x+1} = \dfrac{1}{\sqrt2}\tan^{-1}\left(\dfrac{3x+1}{\sqrt2}\right) [∵ \displaystyle\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=\dfrac{\sqrt2}{3}]
Multiplying by B=-\dfrac{11}{3}: -\dfrac{11}{3\sqrt2}\tan^{-1}\left(\dfrac{3x+1}{\sqrt2}\right) = -\dfrac{11\sqrt2}{6}\tan^{-1}\left(\dfrac{3x+1}{\sqrt2}\right)
(x-5)(x-4)=x^2-9x+20. Express 6x+7 = A(2x-9)+B
Comparing coefficients: 2A=6\Rightarrow A=3, and -9A+B=7\Rightarrow B=34
First part: 3\int \dfrac{2x-9}{\sqrt{x^2-9x+20}}\,dx = 3\cdot2\sqrt{x^2-9x+20} = 6\sqrt{x^2-9x+20} [∵ \displaystyle\int \dfrac{f'}{\sqrt f}\,dx = 2\sqrt f]
Complete the square for the second part: x^2-9x+20 = \left(x-\dfrac92\right)^2-\dfrac14, so \displaystyle\int \dfrac{dx}{\sqrt{x^2-9x+20}} = \log\left|\left(x-\dfrac92\right)+\sqrt{x^2-9x+20}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right|\text{, here }a=\dfrac12]
Multiplying by B=34:
Express x+2 = A(4-2x)+B
Comparing coefficients: -2A=1\Rightarrow A=-\dfrac12, and 4A+B=2\Rightarrow B=4
First part: -\dfrac12\int \dfrac{4-2x}{\sqrt{4x-x^2}}\,dx = -\dfrac12\cdot2\sqrt{4x-x^2} = -\sqrt{4x-x^2} [∵ \displaystyle\int \dfrac{f'}{\sqrt f}\,dx = 2\sqrt f]
Complete the square for the second part: 4x-x^2 = 4-(x-2)^2, so \displaystyle\int \dfrac{dx}{\sqrt{4x-x^2}} = \sin^{-1}\left(\dfrac{x-2}{2}\right) [∵ \displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=2]
Multiplying by B=4:
Express x+2 = A(2x+2)+B
Comparing coefficients: 2A=1\Rightarrow A=\dfrac12, and 2A+B=2\Rightarrow B=1
First part: \dfrac12\int \dfrac{2x+2}{\sqrt{x^2+2x+3}}\,dx = \dfrac12\cdot2\sqrt{x^2+2x+3} = \sqrt{x^2+2x+3} [∵ \displaystyle\int \dfrac{f'}{\sqrt f}\,dx = 2\sqrt f]
Complete the square for the second part: x^2+2x+3 = (x+1)^2+2, so \displaystyle\int \dfrac{dx}{\sqrt{x^2+2x+3}} = \log\left|(x+1)+\sqrt{x^2+2x+3}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|\text{, here }a=\sqrt2]
Multiplying by B=1:
Express x+3 = A(2x-2)+B
Comparing coefficients: 2A=1\Rightarrow A=\dfrac12, and -2A+B=3\Rightarrow B=4
First part: \dfrac12\int \dfrac{2x-2}{x^2-2x-5}\,dx = \dfrac12\log|x^2-2x-5| [∵ \displaystyle\int \dfrac{f'}{f}\,dx = \log|f|]
Complete the square for the second part: x^2-2x-5=(x-1)^2-6, so \displaystyle\int \dfrac{dx}{(x-1)^2-6} = \dfrac{1}{2\sqrt6}\log\left|\dfrac{x-1-\sqrt6}{x-1+\sqrt6}\right| [∵ \displaystyle\int \dfrac{dx}{x^2-a^2} = \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|\text{, here }a=\sqrt6]
Multiplying by B=4: 4\cdot\dfrac{1}{2\sqrt6} = \dfrac{2}{\sqrt6} = \dfrac{\sqrt6}{3}
Express 5x+3 = A(2x+4)+B
Comparing coefficients: 2A=5\Rightarrow A=\dfrac52, and 4A+B=3\Rightarrow B=-7
First part: \dfrac52\int \dfrac{2x+4}{\sqrt{x^2+4x+10}}\,dx = \dfrac52\cdot2\sqrt{x^2+4x+10} = 5\sqrt{x^2+4x+10} [∵ \displaystyle\int \dfrac{f'}{\sqrt f}\,dx = 2\sqrt f]
Complete the square for the second part: x^2+4x+10=(x+2)^2+6, so \displaystyle\int \dfrac{dx}{\sqrt{x^2+4x+10}} = \log\left|(x+2)+\sqrt{x^2+4x+10}\right| [∵ \displaystyle\int \dfrac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|\text{, here }a=\sqrt6]
Multiplying by B=-7:
Complete the square: x^2+2x+2=(x+1)^2+1
Put x+1=t: \displaystyle\int \dfrac{dx}{x^2+2x+2} = \int \dfrac{dt}{t^2+1} = \tan^{-1}t [∵ \displaystyle\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=1]
Factor out the coefficient of x^2 first: 9x-4x^2 = -4\left[x^2-\dfrac94 x\right]
Complete the square inside the bracket: -4\left[x^2-\dfrac94 x+\left(\dfrac98\right)^2-\left(\dfrac98\right)^2\right] = -4\left[\left(x-\dfrac98\right)^2-\left(\dfrac98\right)^2\right] = 4\left[\left(\dfrac98\right)^2-\left(x-\dfrac98\right)^2\right]
So \sqrt{9x-4x^2} = 2\sqrt{\left(\frac98\right)^2-\left(x-\frac98\right)^2}
Put x-\dfrac98=t: \displaystyle\int \dfrac{dx}{\sqrt{9x-4x^2}} = \dfrac12\int \dfrac{dt}{\sqrt{\left(\frac98\right)^2-t^2}} = \dfrac12\sin^{-1}\left(\dfrac{t}{9/8}\right) [∵ \displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right)\text{, here }a=\dfrac98], i.e. \dfrac12\sin^{-1}\left(\dfrac{8x-9}{9}\right)
Expert CBSE Coaching · Class 9–12