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![\large\begin{array}{l} \textsf{Then (iii) becomes}\\\\ \mathsf{\displaystyle\int\!tan^{-1}(4x)\,dx=\frac{1}{4}\,tan^{-1}(4x)\cdot 4x+\frac{1}{4}\,\ell n\left|\dfrac{1}{\sqrt{16x^2+1}}\right|+C}\\\\\\ \textsf{Simplifying...}\\\\ \mathsf{\displaystyle\int\!tan^{-1}(4x)\,dx=\frac{1}{\diagup\!\!\!\! 4}\,tan^{-1}(4x)\cdot \diagup\!\!\!\! 4x+\frac{1}{4}\,\ell n\Big[\left(16x^2+1\right)^{\!\!-\frac{1}{2}}\Big]+C}\\\\ \mathsf{\displaystyle\int\!tan^{-1}(4x)\,dx=x\cdot tan^{-1}(4x)+\frac{1}{4}\cdot \left(-\dfrac{1}{2}\right)\,\ell n\!\left(16x^2+1\right)+C} \end{array}](/tpl/images/0261/9388/95086.png)
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