$ \int\tan^{-1}\sqrt x$ is equal to which of the following options:

Question

$\begin{array}{1 1}(A)\;(x+1)\tan^{-1}\sqrt x-\sqrt x+C & (B)\;x\tan^{-1}\sqrt x-\sqrt x+C\\(C)\;\sqrt x-x\tan^{-1}\sqrt x+C & (D)\;\sqrt x-(x+1)\tan^{-1}\sqrt x+C\end{array}$

Answer 1

Toolbox:

• If $f(x)$ is substituted by $f(t)$ and if $f'(x)dx=f'(t)dt$,then $\int f(x)dx=\int f(t)dt.$
• $\int\large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)+c.$

Step 1:

Let $I=\int\tan^{-1}\sqrt x dx.$

Put $x=t^2$,then on differentiating with respect to $t$,we get $dx=2t.dt$

$I=\int \tan^{-1}t.2t \;dt.$

$\;\;=2\int t\tan^{-1}t.dt.$

Clearly this is of the form $\int u dv.$

Hence this can be solved by the method of integration by parts.

$\int udv=uv-\int vdu.$

Let $u=\tan^{-1}t$

On differentiating we get,

$du=\large\frac{1}{1+t^2}$$dt$

Let $dv=t dt$,on integrating we get,

$v=\large\frac{t^2}{2}$

Now substituting for $u,v,du$ and $dv$,we get

$2\int t.\tan^{-1}t dt=2[\large\frac{t^2}{2}$$\tan^{-1}(t)-\int\large\frac{t^2}{2}.\frac{1}{1+t^2}$$dt$]

Step 2:

Consider $\int\large \frac{t^2}{1+t^2}$$dt$

Add and subtract 1 to the numerator

$\;\;=\int \large\frac{t^2+1-1}{1+t^2}$$dt$.

On splitting the terms as,

$\int\large\frac{t^2+1}{1+t^2}$$dt$$-\int\large\frac{1}{1+t^2}$$dt$

$\;\;=\int dt-\int \large\frac{1}{1+t^2}$$dt$

Step 3:

Now integrating this we get,

$t-\tan^{-1}(t)+c$

Hence $I=2\begin{bmatrix}\large\frac{t^2}{2}\normalsize\tan^{-1}(t)-\large\frac{1}{2}\normalsize(t-\tan^{-1}(t)\end{bmatrix}+c.$

$\qquad\;\;=t^2\tan^{-1}t-t+\tan^{-1}t+c$

Combining the like terms,

$\;\;=\tan^{-1}t(t^2+1)-t+c.$

Substituting for $t$ we get

$I=\tan^{-1}\sqrt x(x+1)-\sqrt x+c.$

Hence the correct option is A.