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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following integrals:
$\int\frac{\text{x}^5}{\sqrt{1+\text{x}^2}}\text{ dx}$
✓

Answer

$\text{I}=\int\frac{\text{x}^5}{\sqrt{1+\text{x}^2}}\text{ dx}\ ....(1)$

Let $1+\text{x}^3=\text{t}^2$ then,

$\text{d}\big(1+\text{x}^3\big)=\text{d}\big(\text{t}^2\big)$

$\Rightarrow3\text{x}^2\text{dx}=\text{dt }2\text{t}$

$\Rightarrow\text{dx}=\frac{\text{dt}}{3\text{x}^2}\text{ 2}\text{t}$

Putting $1+\text{x}^3=\text{t}^2$ and $\text{dx}=\frac{2\text{t}}{3\text{x}^2}\text{ dt}$ in equation (1),

we get,

,$\text{I}=\int\frac{\text{x}^5}{\sqrt{{t}^2}}\times\frac{2\text{t}}{3\text{x}^2}\text{ dt}$

$=\int\frac{\text{x}^5}{\text{t}}\times\frac{2\text{t}}{3\text{x}^2}\text{ dt}$

$=\frac{2}{3}\int\text{x}^3\text{dt}$

$=\frac{2}{3}\int\big(\text{t}^2-1\big)\text{dt}$

$=\frac{2}{3}\times\frac{\text{t}^3}{3}-\frac{2}{3}\text{t}+\text{C}$

$\text{I}=\frac{2}{9}\big(1+\text{x}^3\big)^{\frac{3}{2}}-\frac{2}{3}\sqrt{1+\text{x}^3}+\text{C}$

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