Answer
\(
\frac{4 x^{\frac{3}{4}}}{3}-6 \sqrt{x}+36 \sqrt[4]{x}-108 \ln (\sqrt[4]{x}+3)+C
\)
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.
\[
\frac{4(\sqrt[4]{x})^{3}}{3}-6(\sqrt[4]{x})^{2}+36 \sqrt[4]{x}-108 \ln (\sqrt[4]{x}+3)
\]
(15) Add constant.
\[
\frac{4 x^{\frac{3}{4}}}{3}-6 \sqrt{x}+36 \sqrt[4]{x}-108 \ln (\sqrt[4]{x}+3)+C
\]
