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Indefinite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int(\text{x}+1)\sqrt{2\text{x}^2+3}\text{dx}$
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Answer

Let $\text{I}=\int(\text{x}+1)\sqrt{2\text{x}^2+3}\text{dx}$
Let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(2\text{x}^2+3)+\mu$
$=\lambda(4\text{x})+\mu$
Equating similar terms, we get,
$4\lambda=1\ \Rightarrow\ \lambda=\frac{1}{4}$
$\mu=1$
$\therefore\ \text{I}=\int\frac{1}{4}(4\text{x})\sqrt{2\text{x}^2+3}\text{dx}+\int1.\sqrt{2\text{x}^2+3}\text{dx}$
Let $2\text{x}^2+3=\text{t}$
$\Rightarrow4\text{x dx}=\text{dt}$
$\text{I}=\frac{1}{4}\int\sqrt{\text{t}}\text{dt}+\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}}\text{dx}$
$=\frac{1}{4}.\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\sqrt2\begin{Bmatrix}\frac{\text{x}}{2}\sqrt{\text{x}^2+\frac{3}{2}}\\+\frac{3}{4}\log\Big|\text{x}+\sqrt{\text{x}^2+\frac{3}{2}}\Big|+\text{C}\end{Bmatrix}$
Hence,
$\text{I}=\frac{1}{6}(2\text{x}^2+3)^{\frac{3}{2}}+\frac{\text{x}}{2}\sqrt{2\text{x}^2+3}\\+\frac{3}{2\sqrt2}\log\Big|\text{x}+\sqrt{\text{x}^2+\frac{3}{2}}\Big|+\text{C}$

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