Question

integration of (tan^-1)root(x)

Answer 1

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✔∫ tan^-1 √x dx

Put √x = t

x = t^2

dx = 2tdt.

=∫tan^-1 t . 2tdt

=2∫t tan^-1 dt.

=2[ tan^-1 t ∫  t.dt - ∫ [ $\frac{d}{dx}$ (tan^-1). ∫  t. dt]
dx

=2[ tan^-1 t . $\frac{ t^2}{2}$ - ∫ ($\frac{1}{1 + t^2 }$. $\frac{t^2}{2}$]dx

=2[ $\frac{t^2}{2}$ tan^-1 - $\frac{2}{2}$ ∫ $\frac{t^2}{1 + t^2 }$ dt.

=t^2 tan^-1 t - $\frac{2}{2}$ ∫ $\frac{(t^2 + 1) - 1}{1 +t^2 }$ .dt

=t^2 tan^-1 t - $\frac{ 2}{2}$ [∫dt - ∫ $\frac{1}{1 + t^2}$.dt]

=t^2 tan^-1 t - $\frac{ 2}{2}$ [ (t) - tan^-1 t ]

=t^2 tan^-1 t - t + tan^-1 t + c

=x  tan^-1 √x - √x + tan^-1√x + c

=tan^-1 √x( x + 1 ) - √x + c

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