Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 06/Solution 1

Recall the following integration by parts formula for the indefinite integral

$$\int u\;\mathrm {d} v=uv-\int v\mathrm {d} u.$$

Integrate by parts by setting $u=\arctan(2x)$ and $\mathrm {d} v=\mathrm {d} x$. Then $\mathrm {d} u={\frac {2}{4x^{2}+1}}\mathrm {d} x$, and $v=x$. Substituting these values in the formula gives,

$${\begin{aligned}\int \arctan(2x)\;\mathrm {d} x=x\arctan(2x)-2\int {\frac {x}{4x^{2}+1}}\;\mathrm {d} x.\end{aligned}}$$

For the second integral we can do the following substitution:

$$u=4x^{2}+1,\;\mathrm {d} u=8x\mathrm {d} x.$$

Finally, we have

$${\begin{aligned}\int \arctan(2x)\;\mathrm {d} x&=x\arctan(2x)-{\frac {1}{4}}\int {\frac {1}{u}}\;\mathrm {d} u\\&=x\arctan(2x)-{\frac {1}{4}}\log |4x^{2}+1|+C.\\\end{aligned}}$$

(Note that the expression $4x^{2}+1$ is always positive, so it would also be correct to remove the absolute value and simply write $\log(4x^{2}+1)$ instead of $\log |4x^{2}+1|$.)