Final Answer
$\frac{1}{2}x^2-2x+\frac{1}{x+1}+3\ln\left(x+1\right)+C_0$
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Step-by-step Solution
Specify the solving method
1
Simplify $\left(\sqrt{1+x}\right)^4$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $\frac{1}{2}$ and $n$ equals $4$
$\int\frac{x^3}{\left(1+x\right)^{2}}dx$
2
Expand
$\int\frac{x^3}{1+2x+x^2}dx$
3
Divide $x^3$ by $1+2x+x^2$
$\begin{array}{l}\phantom{\phantom{;}x^{2}+2x\phantom{;}+1;}{\phantom{;}x\phantom{;}-2\phantom{;}\phantom{;}}\\\phantom{;}x^{2}+2x\phantom{;}+1\overline{\smash{)}\phantom{;}x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{\phantom{;}x^{2}+2x\phantom{;}+1;}\underline{-x^{3}-2x^{2}-x\phantom{;}\phantom{-;x^n}}\\\phantom{-x^{3}-2x^{2}-x\phantom{;};}-2x^{2}-x\phantom{;}\phantom{-;x^n}\\\phantom{\phantom{;}x^{2}+2x\phantom{;}+1-;x^n;}\underline{\phantom{;}2x^{2}+4x\phantom{;}+2\phantom{;}\phantom{;}}\\\phantom{;\phantom{;}2x^{2}+4x\phantom{;}+2\phantom{;}\phantom{;}-;x^n;}\phantom{;}3x\phantom{;}+2\phantom{;}\phantom{;}\\\end{array}$
4
Resulting polynomial
$\int\left(x-2+\frac{3}{1+x}+\frac{-1}{\left(1+x\right)^{2}}\right)dx$
5
Simplify the expression inside the integral
$\int xdx+\int-2dx+\int\frac{3}{1+x}dx+\int\frac{-1}{\left(1+x\right)^{2}}dx$
6
The integral $\int xdx$ results in: $\frac{1}{2}x^2$
$\frac{1}{2}x^2$
7
The integral $\int-2dx$ results in: $-2x$
$-2x$
8
The integral $\int\frac{3}{1+x}dx+\int\frac{-1}{\left(1+x\right)^{2}}dx$ results in: $3\ln\left(x+1\right)+\frac{1}{x+1}$
$3\ln\left(x+1\right)+\frac{1}{x+1}$
9
Gather the results of all integrals
$\frac{1}{2}x^2-2x+\frac{1}{x+1}+3\ln\left(x+1\right)$
10
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{1}{2}x^2-2x+\frac{1}{x+1}+3\ln\left(x+1\right)+C_0$
Final Answer
$\frac{1}{2}x^2-2x+\frac{1}{x+1}+3\ln\left(x+1\right)+C_0$