"Solve the following integrals. ∫x2x2+6dx ∫dx(2+x2)32" was answered by students like you

iricky827b

Answered question

2021-11-19

Solve the following integrals.

\int \frac{x^{2}}{\sqrt{x^{2} + 6}} d x

\int \frac{d x}{{(2 + x^{2})}^{\frac{3}{2}}}

Answer & Explanation

Xyle1991

Beginner2021-11-20Added 15 answers

Given integral is

\int \frac{x^{2}}{\sqrt{x^{2} + 6}} d x

Let,

x = \sqrt{6} \tan (u)

d x = \sqrt{6} \sec^{2} (u) d u

and

u = \arctan (\frac{x}{\sqrt{6}})

Then, with this substitution, the integral becomes,

\int \frac{6^{\frac{2}{3}} \sec^{2} (u) \tan^{2} (u)}{\sqrt{6 \tan^{2} (u) + 6}} d u

= 6 \int \sec (u) \tan^{2} (u) d u

= 6 \int \sec (u) [\sec^{2} (u) - 1] d u

= 6 \int \sec^{3} (u) d u - 6 \int \sec (u) d u

= 6 [\frac{\sec (u) \tan (u)}{2} + \frac{1}{2} \sec (u) d u] - 6 \int \sec (u) d u

= 3 \sec (u) \tan (u) - 3 \ln (\tan (u) + \sec (u)) + C

= \frac{3 x \sqrt{\frac{x^{2}}{6} + 1}}{\sqrt{6}} - 3 \ln (\sqrt{\frac{x^{2}}{6} + 1} + \frac{x}{\sqrt{6}})) + C

where C is the integral constant.

Antum1978

Beginner2021-11-21Added 15 answers

The given integral is
$\int \frac{d x}{{(2 + x^{2})}^{\frac{3}{2}}}$
Let,
$x = \sqrt{2} \tan (u)$
$\Rightarrow d x = \sqrt{2} \sec^{2} (u) d u$
and
$u = \arctan (\frac{x}{\sqrt{2}})$
Then, the integral becomes
$\int \frac{\sqrt{2} \sec^{2} (u)}{{(2 \tan^{2} (u) + 2)}^{\frac{3}{2}}} d u$
$= \frac{1}{2} \int \frac{1}{\sec (u)} d u$
$= \frac{1}{2} \int \cos (u) d u$
$= \frac{1}{2} \sin (u) + C$
$= \frac{x}{2^{\frac{3}{2}} \sqrt{\frac{x^{2}}{2}} + 1} + C$
$= \frac{x}{\sqrt{x^{2} + 2}} + C$
where C is the integral constant.

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