Question

$$\int\frac{1}{1+x^{2}}\sinh(\tan^{-1}x)dx$$

Answer

$$(e^2*IM*n*t^2*g*r*a*d*x*sin(h)*arctan(x))/(1+x^2)$$

Solution


Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{\imath ntegrate\times 1\times \sin{h}(\tan^{-1}{(x)})dx}{1+{x}^{2}}\]
Regroup terms.
\[\frac{nttgradx\imath ee\sin{h}\tan^{-1}{(x)}}{1+{x}^{2}}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{n{t}^{2}gradx\imath {e}^{2}\sin{h}\tan^{-1}{(x)}}{1+{x}^{2}}\]
Regroup terms.
\[\frac{{e}^{2}\imath n{t}^{2}gradx\sin{h}\tan^{-1}{(x)}}{1+{x}^{2}}\]