Integration step need some explanation (Calculus) $\displaystyle{\frac15\int_0^1\left[v(1+v^2)^5-v^{11}\right]dv=\frac 15\left[\frac 12\cdot\frac 16(1+v^2)^6-\frac{1}{12}v^{12}\right]_0^1}$
Can someone explain how to get the RHS from LHS. It's the first term after the integral sign that I am confused about.
Thanks. 
 A: $\frac15\int_0^1\left[v(1+v^2)^5-v^{11}\right]dv=\frac15\int_0^1\left[\frac12(1+v^2)^5-\frac12(v^2)^5\right]d(v^2)$
A: The first integral is being done by substitution. Put $u = v^2$, then $du = 2v\,dv$ so
$$\int_0^1 v(1 + v^2)^5 dv = \int_0^1 u^5(du/2) $$
I think you can see the rest.
A: $$v(1+v^2)^5-v^{11}=\left(\frac 12\cdot\frac 16(1+v^2)^6-\frac{1}{12}v^{12}\right)'$$
$$\int v(1+v^2)^5dv-\int v^{11}dv=I_1-I_2$$
$$I_2=v^{12}/12$$
$$I_1=\int v(1+v^2)^5dv$$
$1+v^2=t,vdv=(1/2)dt$
$$I_1=(1/2)\int t^5dt=(1/2)(1/6)t^6$$
A: Remember that integration is linear so:
$$\int x(1+x^2)^5-x^{11} \, \operatorname{d}\!x = \int x(1+x^2)^5 \, \operatorname{d}\!x - \int x^{11} \, \operatorname{d}\!x$$
Hopefully you already know that 
$$\int x^{11} \, \operatorname{d}\!x = \frac{1}{12}x^{12}+C$$
You need to show that 
$$\int x(1+x^2)^5 \, \operatorname{d}\!x = \frac{1}{12}(1+x^2)^6 + C$$
We can perform this integration by substitution. If $u=1+x^2$ then $\operatorname{d}\!u = 2x\,\operatorname{d}\!x$. Hence
$$\int x(1+x^2)^5 \, \operatorname{d}\!x = \int u^5 \cdot \tfrac{1}{2}\,\operatorname{d}\!u = \frac{1}{2}\int u^5 \, \operatorname{d}\!u = \frac{1}{2}\cdot\frac{1}{6}\cdot u^6 + C$$
Recalling that $u=1+x^2$, we may conclude that
$$\int x(1+x^2)^5 \, \operatorname{d}\!x = \frac{1}{2}\cdot \frac{1}{6} \cdot (1+x^2)^6 + C$$
