Q.Evaluate $\displaystyle\int_{0}^{1} 2x\,(x^{2}+1)^{3}\,dx$ using substitution.

The factor $2x$ is the derivative of $x^2+1$, so substitute $u=x^2+1$ and change the limits.

Follow the substitution method of §4.

Choose the substitution. Let $u = x^2 + 1$. Then $du = 2x\,dx$ — and $2x\,dx$ is exactly the remaining factor in the integrand, so the substitution fits cleanly.

Change the limits. When $x = 0$: $u = 0^2 + 1 = 1$. When $x = 1$: $u = 1^2 + 1 = 2$. So the $x$-limits $0 \to 1$ become the $u$-limits $1 \to 2$.

Rewrite and evaluate in $u$.

$\int_{0}^{1} 2x\,(x^2+1)^3\,dx = \int_{1}^{2} u^3\,du = \left[\frac{u^4}{4}\right]_{1}^{2} = \frac{2^4 - 1^4}{4} = \frac{16 - 1}{4} = \frac{15}{4}.$ …

Direct antiderivative without changing variable: $\displaystyle\int 2x(x^2+1)^3\,dx = \frac{(x^2+1)^4}{4}$, so the integral is $\left[\fr …

The classic mistake is to substitute $u = x^2+1$ but keep the original limits $0$ and $1$, computing $\int_0^1 u^3\,du = \tfrac14$ — wrong, because $0$ and $1$ are $x$-values, no …