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Evaluate $\int \frac{1}{\sqrt[4]{x}+3} d x$
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$\frac{4 x^{\frac{3}{4}}}{3}-6 \sqrt{x}+36 \sqrt[4]{x}-108 \ln (\sqrt[4]{x}+3)+C$

Explanation

(1) Use Power Substitution.
Let $u=\sqrt[4]{x}, x=u^{4}$, and $d x=4 u^{3} d u$
(2) Substitute variables from above.
$\int \frac{1}{u+3} \times 4 u^{3} d u$
(3) Use Constant Factor Rule: $\int c f(x) d x=c \int f(x) d x$.
$4 \int \frac{u^{3}}{u+3} d u$
4 Polynomial Division
$4 \int u^{2}-3 u+9-\frac{27}{u+3} d u$
(5) Use Sum Rule: $\int f(x)+g(x) d x=\int f(x) d x+\int g(x) d x$.
$4\left(\int u^{2}-3 u+9 d u-\int \frac{27}{u+3} d u\right)$
(6) Use Power Rule: $\int x^{n} d x=\frac{x^{n+1}}{n+1}+C$.
$4\left(\frac{u^{3}}{3}-\frac{3 u^{2}}{2}+9 u-\int \frac{27}{u+3} d u\right)$

(7) Use Constant Factor Rule: $\int c f(x) d x=c \int f(x) d x$.
$4\left(\frac{u^{3}}{3}-\frac{3 u^{2}}{2}+9 u-27 \int \frac{1}{u+3} d u\right)$
(8) Use Integration by Substitution on $\int \frac{1}{u+3} d u$.
Let $w=u+3$, $d w=d u$
(9) Using $w$ and $d w$ above, rewrite $\int \frac{1}{u+3} d u$.
$\int \frac{1}{w} d w$
(10) The derivative of $\ln x$ is $\frac{1}{x}$.
$\ln w$
(11) Substitute $w=u+3$ back into the original integral.
$\ln (u+3)$
(12) Rewrite the integral with the completed substitution.
$4\left(\frac{u^{3}}{3}-\frac{3 u^{2}}{2}+9 u-27 \ln (u+3)\right)$

(13) Expand.
$4\left(\frac{u^{3}}{3}-\frac{3 u^{2}}{2}+9 u-27 \ln (u+3)\right)$
(14) Substitute $u=\sqrt[4]{x}$ back into the original integral.
$\frac{4(\sqrt[4]{x})^{3}}{3}-6(\sqrt[4]{x})^{2}+36 \sqrt[4]{x}-108 \ln (\sqrt[4]{x}+3)$
$\frac{4 x^{\frac{3}{4}}}{3}-6 \sqrt{x}+36 \sqrt[4]{x}-108 \ln (\sqrt[4]{x}+3)+C$

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