NCERT Solutions · Class 12 Mathematics Integrals

Complete the square, then use \(\displaystyle \int \frac{dt}{t^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{t}{a}+\mathrm{C}\). \[9x^{2}+6x+5=\left(9x^{2}+6x+1\right)+4=(3x+1)^{2}+2^{2}\] Put \(\displaystyle t=3x+1\); then \(\displaystyle dt=3\,dx\), i.e. \(\displaystyle dx=\frac{dt}{3}\) — forgetting this \(\displaystyle \frac{1}{3}\) is the usual slip. \[\int \frac{dx}{9x^{2}+6x+5}=\frac{1}{3}\int \frac{dt}{t^{2}+2^{2}}=\frac{1}{3}\cdot\frac{1}{2}\tan^{-1}\frac{t}{2}+\mathrm{C}\] Hence \[\int \frac{dx}{9x^{2}+6x+5}=\frac{1}{6}\tan^{-1}\!\left(\frac{3x+1}{2}\right)+\mathrm{C}\]

Complete the square. Because the coefficient of \(\displaystyle x^{2}\) is negative, take \(\displaystyle -1\) out first. \[7-6x-x^{2}=7-\left(x^{2}+6x\right)=7-\left[(x+3)^{2}-9\right]=16-(x+3)^{2}=4^{2}-(x+3)^{2}\] Put \(\displaystyle t=x+3\), \(\displaystyle dt=dx\), and use \(\displaystyle \int \frac{dt}{\sqrt{a^{2}-t^{2}}}=\sin^{-1}\frac{t}{a}+\mathrm{C}\). \[\int \frac{dx}{\sqrt{7-6x-x^{2}}}=\int \frac{dt}{\sqrt{4^{2}-t^{2}}}=\sin^{-1}\frac{t}{4}+\mathrm{C}\] Hence \[\int \frac{dx}{\sqrt{7-6x-x^{2}}}=\sin^{-1}\!\left(\frac{x+3}{4}\right)+\mathrm{C},\qquad -7<x<1\]

Expand the product first, then complete the square — the factored form is not itself a standard form. \[(x-1)(x-2)=x^{2}-3x+2=\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}+2=\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\] Put \(\displaystyle t=x-\frac{3}{2}\), \(\displaystyle dt=dx\), and use \(\displaystyle \int \frac{dt}{\sqrt{t^{2}-a^{2}}}=\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}\). \[\int \frac{dx}{\sqrt{(x-1)(x-2)}}=\int \frac{dt}{\sqrt{t^{2}-\left(\dfrac{1}{2}\right)^{2}}}=\log\left|t+\sqrt{t^{2}-\tfrac{1}{4}}\right|+\mathrm{C}\] Hence \[\int \frac{dx}{\sqrt{(x-1)(x-2)}}=\log\left|\left(x-\frac{3}{2}\right)+\sqrt{x^{2}-3x+2}\right|+\mathrm{C}\] valid where the radicand is positive, i.e. for \(\displaystyle x>2\) or \(\displaystyle x<1\).

Complete the square, taking \(\displaystyle -1\) out of the \(\displaystyle x\)-terms because the \(\displaystyle x^{2}\) coefficient is negative. \[8+3x-x^{2}=8-\left(x^{2}-3x\right)=8-\left[\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}\right]=\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}\] So the radicand is \(\displaystyle \left(\frac{\sqrt{41}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}\). Put \(\displaystyle t=x-\frac{3}{2}\), \(\displaystyle dt=dx\), and use \(\displaystyle \int \frac{dt}{\sqrt{a^{2}-t^{2}}}=\sin^{-1}\frac{t}{a}+\mathrm{C}\). \[\int \frac{dx}{\sqrt{8+3x-x^{2}}}=\sin^{-1}\!\left(\frac{x-\dfrac{3}{2}}{\dfrac{\sqrt{41}}{2}}\right)+\mathrm{C}=\sin^{-1}\!\left(\frac{2x-3}{\sqrt{41}}\right)+\mathrm{C}\]

Multiply out and complete the square; the half-sum \(\displaystyle \frac{a+b}{2}\) is the shift and the half-difference \(\displaystyle \frac{a-b}{2}\) is the constant. \[(x-a)(x-b)=x^{2}-(a+b)x+ab=\left(x-\frac{a+b}{2}\right)^{2}-\frac{(a+b)^{2}}{4}+ab\] and \(\displaystyle \dfrac{(a+b)^{2}}{4}-ab=\dfrac{(a-b)^{2}}{4}\), so \[(x-a)(x-b)=\left(x-\frac{a+b}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}\] Put \(\displaystyle t=x-\frac{a+b}{2}\), \(\displaystyle dt=dx\), and use \(\displaystyle \int \frac{dt}{\sqrt{t^{2}-c^{2}}}=\log\left|t+\sqrt{t^{2}-c^{2}}\right|+\mathrm{C}\). \[\int \frac{dx}{\sqrt{(x-a)(x-b)}}=\log\left|\left(x-\frac{a+b}{2}\right)+\sqrt{(x-a)(x-b)}\right|+\mathrm{C}\] valid on the intervals where \(\displaystyle (x-a)(x-b)>0\).

Check the numerator against the derivative of the radicand before doing anything else. \[\frac{d}{dx}\left(2x^{2}+x-3\right)=4x+1,\] which is exactly the numerator. So put \(\displaystyle u=2x^{2}+x-3\), giving \(\displaystyle du=(4x+1)\,dx\). \[\int \frac{4x+1}{\sqrt{2x^{2}+x-3}}\,dx=\int \frac{du}{\sqrt{u}}=\int u^{-1/2}\,du=2\sqrt{u}+\mathrm{C}\] Hence \[\int \frac{4x+1}{\sqrt{2x^{2}+x-3}}\,dx=2\sqrt{2x^{2}+x-3}+\mathrm{C}\]

Split the numerator into a multiple of \(\displaystyle \frac{d}{dx}\left(x^{2}-1\right)=2x\) plus a constant: \[x+2=\frac{1}{2}(2x)+2\] \[\int \frac{x+2}{\sqrt{x^{2}-1}}\,dx=\frac{1}{2}\int \frac{2x}{\sqrt{x^{2}-1}}\,dx+2\int \frac{dx}{\sqrt{x^{2}-1}}\] For the first, \(\displaystyle u=x^{2}-1\), \(\displaystyle du=2x\,dx\), so \(\displaystyle \frac{1}{2}\int u^{-1/2}du=\sqrt{x^{2}-1}\). The second is the standard form \(\displaystyle \int \frac{dx}{\sqrt{x^{2}-a^{2}}}=\log\left|x+\sqrt{x^{2}-a^{2}}\right|\) with \(\displaystyle a=1\).Therefore \[\int \frac{x+2}{\sqrt{x^{2}-1}}\,dx=\sqrt{x^{2}-1}+2\log\left|x+\sqrt{x^{2}-1}\right|+\mathrm{C}\]

Write the numerator as \(\displaystyle \mathrm{A}\times\)(derivative of the denominator)\(\displaystyle +\mathrm{B}\).Here \(\displaystyle \frac{d}{dx}\left(3x^{2}+2x+1\right)=6x+2\), so let \[5x-2=\mathrm{A}(6x+2)+\mathrm{B}\] Comparing coefficients of \(\displaystyle x\): \(\displaystyle 6\mathrm{A}=5\Rightarrow \mathrm{A}=\frac{5}{6}\). Comparing constants: \(\displaystyle 2\mathrm{A}+\mathrm{B}=-2\Rightarrow \mathrm{B}=-2-\frac{5}{3}=-\frac{11}{3}\). \[\int \frac{5x-2}{3x^{2}+2x+1}\,dx=\frac{5}{6}\int \frac{6x+2}{3x^{2}+2x+1}\,dx-\frac{11}{3}\int \frac{dx}{3x^{2}+2x+1}\] The first integral is \(\displaystyle \frac{5}{6}\log\left(3x^{2}+2x+1\right)\) (the quadratic has discriminant \(\displaystyle 4-12<0\), so it is always positive and no modulus is needed).For the second, complete the square after taking out the leading \(\displaystyle 3\): \[3x^{2}+2x+1=3\left[\left(x+\frac{1}{3}\right)^{2}+\frac{2}{9}\right]\] \[\int \frac{dx}{3x^{2}+2x+1}=\frac{1}{3}\int \frac{dx}{\left(x+\dfrac{1}{3}\right)^{2}+\left(\dfrac{\sqrt{2}}{3}\right)^{2}}=\frac{1}{3}\cdot\frac{3}{\sqrt{2}}\tan^{-1}\!\left(\frac{3x+1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}\tan^{-1}\!\left(\frac{3x+1}{\sqrt{2}}\right)\] Therefore \[\int \frac{5x-2}{1+2x+3x^{2}}\,dx=\frac{5}{6}\log\left(3x^{2}+2x+1\right)-\frac{11}{3\sqrt{2}}\tan^{-1}\!\left(\frac{3x+1}{\sqrt{2}}\right)+\mathrm{C}\]

First expand the product: \(\displaystyle (x-5)(x-4)=x^{2}-9x+20\), whose derivative is \(\displaystyle 2x-9\).Write \(\displaystyle 6x+7=\mathrm{A}(2x-9)+\mathrm{B}\). Coefficients of \(\displaystyle x\): \(\displaystyle 2\mathrm{A}=6\Rightarrow \mathrm{A}=3\). Constants: \(\displaystyle -9\mathrm{A}+\mathrm{B}=7\Rightarrow \mathrm{B}=7+27=34\). \[\int \frac{6x+7}{\sqrt{x^{2}-9x+20}}\,dx=3\int \frac{2x-9}{\sqrt{x^{2}-9x+20}}\,dx+34\int \frac{dx}{\sqrt{x^{2}-9x+20}}\] For the first, \(\displaystyle u=x^{2}-9x+20\), \(\displaystyle du=(2x-9)\,dx\), giving \(\displaystyle 3\int u^{-1/2}du=6\sqrt{x^{2}-9x+20}\).For the second, complete the square: \[x^{2}-9x+20=\left(x-\frac{9}{2}\right)^{2}-\frac{81}{4}+20=\left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\] so \(\displaystyle \int \frac{dx}{\sqrt{x^{2}-9x+20}}=\log\left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9x+20}\right|\).Therefore \[\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx=6\sqrt{x^{2}-9x+20}+34\log\left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9x+20}\right|+\mathrm{C}\]

The derivative of \(\displaystyle 4x-x^{2}\) is \(\displaystyle 4-2x\), so split the numerator against that.Let \(\displaystyle x+2=\mathrm{A}(4-2x)+\mathrm{B}\). Coefficients of \(\displaystyle x\): \(\displaystyle -2\mathrm{A}=1\Rightarrow \mathrm{A}=-\frac{1}{2}\). Constants: \(\displaystyle 4\mathrm{A}+\mathrm{B}=2\Rightarrow \mathrm{B}=4\). \[\int \frac{x+2}{\sqrt{4x-x^{2}}}\,dx=-\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^{2}}}\,dx+4\int \frac{dx}{\sqrt{4x-x^{2}}}\] First part: with \(\displaystyle u=4x-x^{2}\), \(\displaystyle du=(4-2x)\,dx\), so \(\displaystyle -\frac{1}{2}\int u^{-1/2}du=-\sqrt{4x-x^{2}}\).Second part: complete the square, \(\displaystyle 4x-x^{2}=4-(x-2)^{2}=2^{2}-(x-2)^{2}\), so \[4\int \frac{dx}{\sqrt{2^{2}-(x-2)^{2}}}=4\sin^{-1}\!\left(\frac{x-2}{2}\right)\] Therefore \[\int \frac{x+2}{\sqrt{4x-x^{2}}}\,dx=-\sqrt{4x-x^{2}}+4\sin^{-1}\!\left(\frac{x-2}{2}\right)+\mathrm{C},\qquad 0<x<4\]