question evaluate the integral below. ∫tan⁻¹(4x)dx be sure to place the argument of any inverse…

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \tan^{- 1}(4x)$ and $dv=dx$. Then $du=\frac{4}{1 + 16x^{2}}dx$ and $v=x$.

Step2: Apply the formula

$\int\tan^{-1}(4x)dx=x\tan^{-1}(4x)-\int\frac{4x}{1 + 16x^{2}}dx$.

Step3: Solve the new integral

Let $t = 1+16x^{2}$, then $dt = 32x;dx$ and $x;dx=\frac{1}{32}dt$. So $\int\frac{4x}{1 + 16x^{2}}dx=\frac{4}{32}\int\frac{dt}{t}=\frac{1}{8}\ln|t|+C=\frac{1}{8}\ln(1 + 16x^{2})+C$.

Answer:

$x\tan^{-1}(4x)-\frac{1}{8}\ln(1 + 16x^{2})+C$