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?

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 $\int u \mathrm dv=vu-\int v \mathrm du$.
