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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$
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Answer

$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$ Let $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Now, $\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}+\text{a}^2}+\sqrt{\text{t}-\text{a}^2}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\Big(\sqrt{\text{t}+\text{a}^2}+\sqrt{\text{t}-\text{a}^2}\Big)}\times\frac{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}$
$=\frac{1}{2}\int\frac{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}{(\text{t}+\text{a}^2)-(\text{t}-\text{a}^2)}\text{ dt}$
$=\frac{1}{4\text{a}^2}\int\Big(\text{t}+\text{a}^2\Big)^{\frac12}\text{dt}-\frac{1}{4\text{a}^2}\big(\text{t}-\text{a}^2\big)^{\frac12}\text{dt}$
$=\frac{1}{4\text{a}^2}\begin{bmatrix}\frac{\big(\text{t}+\text{a}^2\big)^{\frac12+1}}{\frac{1}2+1}\end{bmatrix}-\frac{1}{4\text{a}^2}\begin{bmatrix}\frac{\big(\text{t}-\text{a}^2\big)^{\frac12+1}}{\frac12+1}\end{bmatrix}+\text{C}$
$=\frac{1}{6\text{a}^2}\begin{bmatrix}(\text{t}+\text{a}^2)^{\frac{3}{2}}-(\text{t}-\text{a}^2)^{\frac32}\end{bmatrix}+\text{C}$
$=\frac{1}{6\text{a}^2}\begin{bmatrix}(\text{x}^2+\text{a}^2)^{\frac32}-(\text{x}^2-\text{a}^2)^{\frac32}\end{bmatrix}+\text{C}$

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