⇒∫ x^1/3(2+x^2/3)^1/4dx Substitute u=x^2/3+2→ dx=√(x)2dx =32∫ (4^5/4 2(u)^1/4)du =32[∫ u^5/4du 2∫ (u)^1/4du] =32[4u^9/49 2[4u^5/45]] ...

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Byju's Answer

Standard XII

Mathematics

nth Term of A.P

∫ x1/3 2 + x2...

Question

$\int x^{1 / 3} (2 + x^{2 / 3})^{1 / 4} d x$

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Solution

$\Rightarrow \int x^{1 / 3} (2 + x^{2 / 3})^{1 / 4} d x$
Substitute $u = x^{2 / 3} + 2 \to d x = \frac{\sqrt[33]{x}}{2} d x$
$= \frac{3}{2} \int (4^{5 / 4} - 2 (u)^{1 / 4}) d u$
$= \frac{3}{2} [\int u^{5 / 4} d u - 2 \int (u)^{1 / 4} d u]$
$= \frac{3}{2} [\frac{4 u^{9 / 4}}{9} - 2 [\frac{4 u^{5 / 4}}{5}]]$
$= \frac{3}{2} [\frac{4 u^{9 / 4}}{9} - \frac{8 u^{5 / 4}}{5}]$
$= \frac{2 u^{9 / 4}}{3} - \frac{12 u^{5 / 4}}{5} + c$
$= \frac{2 {(x^{2 / 3} + 2)}^{9 / 4}}{3} - \frac{12 {(x^{2 / 3} + 2)}^{5 / 4}}{5} + c$ .

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∫x1/3(2+x2/3)1/4dx

⇒∫x1/3(2+x2/3)1/4dxSubstitute u=x2/3+2→dx=33√x2dx=32∫(45/4−2(u)1/4)du=32[∫u5/4du−2∫(u)1/4du]=32[4u9/49−2[4u5/45]]=32[4u9/49−8u5/45]=2u9/43−12u5/45+c=2(x2/3+2)9/43−12(x2/3+2)5/45+c.

$\int x^{1 / 3} (2 + x^{2 / 3})^{1 / 4} d x$