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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals3 Marks
Question
Evaluate the following integrals:
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+1}}\text{ dx}$
✓

Answer

We have
$\text{I}=\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+1}}\text{ dx}$
Putting $\text{x}-1=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\big(\frac{1}{\text{t}}\big)\sqrt{\big(1+\frac{1}{\text{t}}}\big)^2+1}$
$=\int\frac{-\frac{1}{\text{t}}\text{ dt}}{\sqrt{1+\frac{1}{\text{t}^2}+\frac{2}{\text{t}}+1}}$
$=\int\frac{-\frac{1}{\text{t}}\text{ dt}}{\sqrt{\frac{\text{t}^2+1+2\text{t}+\text{t}^2}{\text{t}}}}$
$=\int\frac{-\text{dt}}{\sqrt{2\text{t}^2+2\text{t}+1}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\frac{1}{2}}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}-\frac{1}{4}+\frac{1}{2}}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\big(\text{t}+\frac{1}{2}\big)^2+\big(\frac{1}{2}\big)^2}$
$=-\frac{1}{\sqrt{2}}\log\begin{vmatrix}\text{t}+\frac{1}{2}+\sqrt{\Big(\text{t}+\frac{1}{2}\Big)^2+\frac{1}{4}}\end{vmatrix}+\text{C}$ where $\text{t}=\frac{1}{\text{x}-1}$

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