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

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

Follow the substitution method of §4, adjusting for the constant factor.

Choose the substitution. Let $u = x^2 + 1$. Then $du = 2x\,dx$, so $x\,dx = \tfrac12\,du$ — the numerator $x\,dx$ becomes $\tfrac12\,du$.

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

Rewrite and evaluate in $u$.

Numerically $\tfrac12\log 2 \approx \tfrac12(0.6931) = 0.3466$. …

Direct antiderivative: $\displaystyle\int \frac{x}{x^2+1}\,dx = \frac12\log(x^2+1)$, so the integral is $\left[\ …

The key error is dropping the $\tfrac12$: since $du = 2x\,dx$, the numerator $x\,dx$ equals $\tfrac12\,du$, not $du$. Forgetting this doubles the answer. Als …