<s>[INST] Integarate $\frac{1}{2}f'(x)$ wrt $x^4$ where $f(x)=\tan^{-1}x+\ln\sqrt{1+x}-\ln\sqrt{1-x}$ Integarate $\frac{1}{2}f'(x)$ wrt $x^4$ where $f(x)=\tan^{-1}x+\ln\sqrt{1+x}-\ln\sqrt{1-x}$
I tried: $\int \frac{1}{2}f'(x) \, d(x^4)=\int \frac{1}{2}f'(x)\cdot 4x^3 \, dx$. I integrated it by parts but it goes lengthy and could not get answer.Please help..
 [/INST] We are given $f(x)=\arctan(x)+\frac12 \log(1+x)-\frac12 \log(1-x)$ and are asked to find $F(x)$ where $F(x)$ is given by
$$F(x)=\int \frac12 f'(x)4x^3\,dx \tag 1$$
We can proceed directly by first determining the derivative of $f$.  To that end, we find that 
$$f'(x)=\frac{1}{1+x^2}+\frac12\left(\frac{1}{1+x}+\frac{1}{1-x}\right) \tag 2$$
Next, we can avoid a series of integration by parts steps by making use of the following $3$ identities for $x^3$.
$$\begin{align}
x^3&=x(x^2+1)-x \tag 3\\\\
x^3&=(x^2-x+1)(1+x)-1 \tag 4\\\\
x^3&=-(x^2+x+1)(1-x)+ 1\tag 5
\end{align}$$
Then, using $(2)-(5)$ in $(1)$ yields
$$\begin{align}
F(x)&=2\int  \frac{x(x^2+1)-x}{1+x^2} \,dx \\\\
&+\int \frac{(x^2-x+1)(1+x)-1}{1+x} \,dx \\\\
&+\int \frac{-(x^2+x+1)(1-x)+ 1}{1-x}\,dx\\\\
&=x^2-\log(1+x^2)\\\\
&+\frac13x^3-\frac12x^2+x-\log|1+x|\\\\
&-\frac13x^3-\frac12x^2-x-\log|1-x|\\\\
&-\log|1-x^4|+C
\end{align}$$
Thus, we have 
$$\bbox[5px,border:2px solid #C0A000]{\int \frac12 f'(x)d(x^4)=-\log|1-x^4|+C}$$
</s>