Definite integral of $4x(1+x)^3$ I'm trying to integrate by parts, and have done:$$\int^1_0{4x(1+x)^3}=4\int^1_0{x(1+x)^3}= 4\bigg[\frac{x(1+x)^4}4-\frac{(1+x)^4 }4\bigg]=4\bigg[\frac{(x-1)(1+x)^4}4\bigg]^1_0=
\\ \bigg[(x-1)(1+x)^4 \bigg]^1_0 = 1$$
Why is this wrong?
 A: Look closely at the section $4\left[\frac{x(1+x)^4}{4}-\frac{(1+x)^4}{4}\right]$. One way the integration by parts formula is commonly expressed is $\int u\cdot dv=uv-\int v\cdot du$. It seems like you instead have something like $uv-v\cdot du$ (not technically accurate to say, but it highlights your error), at least if you had let $u=x$ and $dv=(1+x)^3dx$.
A: Integration by parts for definite integrals

$$\int_a^b u\ \mathrm dv=uv\bigg|_a^b-\int_a^b v\ \mathrm du$$

The integration by parts formula wasn't correctly applied in your attempt. Let's take a look at this integral
$$\int_0^1 4x(x+1)^3\ \mathrm dx$$
Applying integration by parts, we have
$$u=x\Rightarrow \mathrm du=\mathrm dx$$
$$v=4\int (x+1)^3\ \mathrm dx=(x+1)^4+C$$
So now
$$x(x+1)^4\bigg|_0^1-\int_0^1 (x+1)^4\ \mathrm dx$$
Simplify the left term and apply integration by substitution 
$$16-\int_1^2 t^4\ \mathrm dt$$
I'll leave the rest to you. 
A: Integration by parts should give, if we take $dv=(1+x)^3 dx$ so that $v=\frac{(1+x)^4}{4}$. And $u=x$ so that $du=dx$, 
$$\int {4x(1+x)^3}=4\int {x(1+x)^3}= 4\bigg[\frac{x(1+x)^4}4-\int \frac{(1+x)^4 }4 \mathrm dx\bigg]$$
As $\int u \mathrm dv=vu-\int v \mathrm du$.
