NCERT Solutions · Class 12 Mathematics Integrals

Split the numerator against \(\displaystyle \frac{d}{dx}\left(x^{2}+2x+3\right)=2x+2\): \[x+2=\frac{1}{2}(2x+2)+1\] \[\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\,dx=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^{2}+2x+3}}\,dx+\int \frac{dx}{\sqrt{x^{2}+2x+3}}\] First part: with \(\displaystyle u=x^{2}+2x+3\), \(\displaystyle du=(2x+2)\,dx\), it is \(\displaystyle \frac{1}{2}\cdot 2\sqrt{u}=\sqrt{x^{2}+2x+3}\).Second part: complete the square, \(\displaystyle x^{2}+2x+3=(x+1)^{2}+\left(\sqrt{2}\right)^{2}\), so \[\int \frac{dx}{\sqrt{(x+1)^{2}+2}}=\log\left|(x+1)+\sqrt{x^{2}+2x+3}\right|\] Therefore \[\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\,dx=\sqrt{x^{2}+2x+3}+\log\left|(x+1)+\sqrt{x^{2}+2x+3}\right|+\mathrm{C}\]

Split the numerator against \(\displaystyle \frac{d}{dx}\left(x^{2}-2x-5\right)=2x-2\): \[x+3=\frac{1}{2}(2x-2)+4\] \[\int \frac{x+3}{x^{2}-2x-5}\,dx=\frac{1}{2}\int \frac{2x-2}{x^{2}-2x-5}\,dx+4\int \frac{dx}{x^{2}-2x-5}\] First part: \(\displaystyle \frac{1}{2}\log\left|x^{2}-2x-5\right|\).Second part: complete the square. The constant is \(\displaystyle -6\), i.e. negative, so this is the \(\displaystyle t^{2}-a^{2}\) form, not an arctangent: \[x^{2}-2x-5=(x-1)^{2}-6=(x-1)^{2}-\left(\sqrt{6}\right)^{2}\] Using \(\displaystyle \int \frac{dt}{t^{2}-a^{2}}=\frac{1}{2a}\log\left|\frac{t-a}{t+a}\right|+\mathrm{C}\), \[4\int \frac{dx}{(x-1)^{2}-6}=4\cdot\frac{1}{2\sqrt{6}}\log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|=\frac{2}{\sqrt{6}}\log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|\] Therefore \[\int \frac{x+3}{x^{2}-2x-5}\,dx=\frac{1}{2}\log\left|x^{2}-2x-5\right|+\frac{2}{\sqrt{6}}\log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+\mathrm{C}\]

Split the numerator against \(\displaystyle \frac{d}{dx}\left(x^{2}+4x+10\right)=2x+4\).Let \(\displaystyle 5x+3=\mathrm{A}(2x+4)+\mathrm{B}\). Coefficients of \(\displaystyle x\): \(\displaystyle 2\mathrm{A}=5\Rightarrow \mathrm{A}=\frac{5}{2}\). Constants: \(\displaystyle 4\mathrm{A}+\mathrm{B}=3\Rightarrow \mathrm{B}=3-10=-7\). \[\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\,dx=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^{2}+4x+10}}\,dx-7\int \frac{dx}{\sqrt{x^{2}+4x+10}}\] First part: with \(\displaystyle u=x^{2}+4x+10\), \(\displaystyle du=(2x+4)\,dx\), it is \(\displaystyle \frac{5}{2}\cdot 2\sqrt{u}=5\sqrt{x^{2}+4x+10}\).Second part: \(\displaystyle x^{2}+4x+10=(x+2)^{2}+6=(x+2)^{2}+\left(\sqrt{6}\right)^{2}\), so \[\int \frac{dx}{\sqrt{(x+2)^{2}+6}}=\log\left|(x+2)+\sqrt{x^{2}+4x+10}\right|\] Therefore \[\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\,dx=5\sqrt{x^{2}+4x+10}-7\log\left|(x+2)+\sqrt{x^{2}+4x+10}\right|+\mathrm{C}\]

Complete the square in the denominator: \[x^{2}+2x+2=(x+1)^{2}+1\] Put \(\displaystyle t=x+1\), so \(\displaystyle dt=dx\), and use \(\displaystyle \int \frac{dt}{t^{2}+1}=\tan^{-1}t+\mathrm{C}\). \[\int \frac{dx}{x^{2}+2x+2}=\int \frac{dt}{t^{2}+1}=\tan^{-1}t+\mathrm{C}=\tan^{-1}(x+1)+\mathrm{C}\] The correct answer is (B) \(\displaystyle \tan^{-1}(x+1)+\mathrm{C}\).

Complete the square, remembering to factor out the coefficient \(\displaystyle 4\) of \(\displaystyle x^{2}\) first — that factor is what produces the \(\displaystyle \frac{1}{2}\) outside. \[9x-4x^{2}=-4\left(x^{2}-\frac{9}{4}x\right)=-4\left[\left(x-\frac{9}{8}\right)^{2}-\frac{81}{64}\right]=4\left[\left(\frac{9}{8}\right)^{2}-\left(x-\frac{9}{8}\right)^{2}\right]\] So \(\displaystyle \sqrt{9x-4x^{2}}=2\sqrt{\left(\frac{9}{8}\right)^{2}-\left(x-\frac{9}{8}\right)^{2}}\), and with \(\displaystyle t=x-\frac{9}{8}\), \(\displaystyle dt=dx\), \[\int \frac{dx}{\sqrt{9x-4x^{2}}}=\frac{1}{2}\int \frac{dt}{\sqrt{\left(\dfrac{9}{8}\right)^{2}-t^{2}}}=\frac{1}{2}\sin^{-1}\!\left(\frac{t}{9/8}\right)+\mathrm{C}\] \[=\frac{1}{2}\sin^{-1}\!\left(\frac{x-\dfrac{9}{8}}{\dfrac{9}{8}}\right)+\mathrm{C}=\frac{1}{2}\sin^{-1}\!\left(\frac{8x-9}{9}\right)+\mathrm{C}\] The correct answer is (B) \(\displaystyle \frac{1}{2}\sin^{-1}\!\left(\frac{8x-9}{9}\right)+\mathrm{C}\).