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.
Write the denominator as (x^3)^2+1 and put t=x^3, so dt=3x^2\,dx.
The integral becomes \displaystyle\int \dfrac{dt}{t^2+1} = \tan^{-1}t.
Write the denominator as \sqrt{1+(2x)^2} and put t=2x, so dt=2\,dx.
The integral becomes \displaystyle\dfrac12\int \dfrac{dt}{\sqrt{t^2+1}} = \dfrac12\log\left|t+\sqrt{t^2+1}\right|.
Put t=2-x, so dt=-dx.
The integral becomes \displaystyle-\int \dfrac{dt}{\sqrt{t^2+1}} = -\log\left|t+\sqrt{t^2+1}\right|.
Write the denominator as \sqrt{3^2-(5x)^2} and put t=5x, so dt=5\,dx.
The integral becomes \displaystyle\dfrac15\int \dfrac{dt}{\sqrt{3^2-t^2}} = \dfrac15\sin^{-1}\left(\dfrac{t}{3}\right).
Put t=x^2, so dt=2x\,dx, i.e. 3x\,dx=\dfrac32\,dt. The denominator becomes 1+2t^2.
The integral becomes \displaystyle\dfrac32\int \dfrac{dt}{1+2t^2} = \dfrac32\cdot\dfrac12\int \dfrac{dt}{t^2+\frac12} = \dfrac34\cdot\sqrt{2}\,\tan^{-1}(\sqrt2\,t).
Write the denominator as 1-(x^3)^2 and put t=x^3, so dt=3x^2\,dx.
The integral becomes \displaystyle\dfrac13\int \dfrac{dt}{1-t^2} = \dfrac13\cdot\dfrac12\log\left|\dfrac{1+t}{1-t}\right|.
Split the fraction: \dfrac{x-1}{\sqrt{x^2-1}} = \dfrac{x}{\sqrt{x^2-1}} - \dfrac{1}{\sqrt{x^2-1}}.
For the first term, put u=x^2-1, du=2x\,dx: \displaystyle\int \dfrac{x}{\sqrt{x^2-1}}\,dx = \sqrt{x^2-1}.
For the second term, use the standard formula \displaystyle\int \dfrac{dx}{\sqrt{x^2-1}} = \log\left|x+\sqrt{x^2-1}\right| (subtracted).
Write a^6=(a^3)^2 and put t=x^3, so dt=3x^2\,dx.
The integral becomes \displaystyle\dfrac13\int \dfrac{dt}{\sqrt{t^2+(a^3)^2}} = \dfrac13\log\left|t+\sqrt{t^2+a^6}\right|.
Put t=\tan x, so dt=\sec^2x\,dx.
The integral becomes \displaystyle\int \dfrac{dt}{\sqrt{t^2+4}} = \log\left|t+\sqrt{t^2+4}\right|.
Complete the square: x^2+2x+2 = (x+1)^2+1.
Put t=x+1: \displaystyle\int \dfrac{dt}{\sqrt{t^2+1}} = \log\left|t+\sqrt{t^2+1}\right|.
Complete the square: 9x^2+6x+5 = 9\left(x+\dfrac13\right)^2+4.
Put t=x+\dfrac13: \displaystyle\int \dfrac{dt}{9t^2+4} = \dfrac19\int \dfrac{dt}{t^2+\frac49} = \dfrac16\tan^{-1}\left(\dfrac{3t}{2}\right).
Complete the square: 7-6x-x^2 = 16-(x+3)^2.
Put t=x+3: \displaystyle\int \dfrac{dt}{\sqrt{16-t^2}} = \sin^{-1}\left(\dfrac{t}{4}\right).
Expand and complete the square: (x-1)(x-2)=x^2-3x+2 = \left(x-\dfrac32\right)^2-\dfrac14.
Put t=x-\dfrac32: \displaystyle\int \dfrac{dt}{\sqrt{t^2-\frac14}} = \log\left|t+\sqrt{t^2-\frac14}\right|.
Complete the square: 8+3x-x^2 = \dfrac{41}{4}-\left(x-\dfrac32\right)^2.
Put t=x-\dfrac32: \displaystyle\int \dfrac{dt}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-t^2}} = \sin^{-1}\left(\dfrac{2t}{\sqrt{41}}\right).
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 t=x-\dfrac{a+b}{2}: \displaystyle\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|.
Notice that \dfrac{d}{dx}(2x^2+x-3) = 4x+1 — exactly the numerator.
Put t=2x^2+x-3: \displaystyle\int \dfrac{dt}{\sqrt{t}} = 2\sqrt{t}.
Express x+2 = A\cdot\dfrac{d}{dx}(x^2-1)+B = A(2x)+B. Comparing coefficients: A=\dfrac12, B=2.
So \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}}.
Evaluate each: \dfrac12\cdot2\sqrt{x^2-1} = \sqrt{x^2-1}, and 2\log\left|x+\sqrt{x^2-1}\right|.
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}.
The first part gives \dfrac56\log(3x^2+2x+1).
For the second part, complete the square: 3x^2+2x+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).
Multiply 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).
Write (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.
The first part gives 3\cdot2\sqrt{x^2-9x+20} = 6\sqrt{x^2-9x+20}.
For the second part, complete the square: 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|.
Multiply by B=34.
Express x+2 = A(4-2x)+B. Comparing coefficients: -2A=1\Rightarrow A=-\dfrac12, and 4A+B=2\Rightarrow B=4.
The first part gives -\dfrac12\cdot2\sqrt{4x-x^2} = -\sqrt{4x-x^2}.
For the second part, complete the square: 4x-x^2 = 4-(x-2)^2, so \displaystyle\int \dfrac{dx}{\sqrt{4x-x^2}} = \sin^{-1}\left(\dfrac{x-2}{2}\right).
Multiply by B=4.
Express x+2 = A(2x+2)+B. Comparing coefficients: 2A=1\Rightarrow A=\dfrac12, and 2A+B=2\Rightarrow B=1.
The first part gives \dfrac12\cdot2\sqrt{x^2+2x+3} = \sqrt{x^2+2x+3}.
For the second part, complete the square: 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|.
Express x+3 = A(2x-2)+B. Comparing coefficients: 2A=1\Rightarrow A=\dfrac12, and -2A+B=3\Rightarrow B=4.
The first part gives \dfrac12\log|x^2-2x-5|.
For the second part, complete the square: 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|.
Multiply 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.
The first part gives \dfrac52\cdot2\sqrt{x^2+4x+10} = 5\sqrt{x^2+4x+10}.
For the second part, complete the square: 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|.
Multiply by B=-7.
Complete the square: x^2+2x+2=(x+1)^2+1.
Put t=x+1: \displaystyle\int \dfrac{dt}{t^2+1} = \tan^{-1}t.
Complete the square: 9x-4x^2 = 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}.
The integral becomes \displaystyle\dfrac12\int \dfrac{dt}{\sqrt{\left(\frac98\right)^2-t^2}} = \dfrac12\sin^{-1}\left(\dfrac{t}{9/8}\right) where t=x-\dfrac98, i.e. \dfrac12\sin^{-1}\left(\dfrac{8x-9}{9}\right).
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