Integration Strategy — Question 2

Question 2 – Solution

Step 1: Choose substitution. The expression $1+x^2$ is inside the logarithm, and its derivative $2x$ matches the remaining factor.

Let $$u=1+x^2,\qquad du=2x\,dx,\qquad x\,dx=\frac12du.$$ Then $$I=\frac12\int\ln u\,du.$$ Step 2: Integrate $\ln u$ by parts. Choose $$v=\ln u,\qquad dw=du.$$ Then $dv=du/u$ and $w=u$, so $$\begin{align*} \int\ln u\,du &=u\ln u-\int u\left(\frac1u\right)du\\ &=u\ln u-u+C. \end{align*}$$ Step 3: Restore the factor $1/2$ and return to $x$. $$\begin{align*} I&=\frac12(u\ln u-u)+C\\ &=\frac12\bigl[(1+x^2)\ln(1+x^2)-(1+x^2)\bigr]+C. \end{align*}$$ $$\boxed{\frac12[(1+x^2)\ln(1+x^2)-(1+x^2)]+C}$$