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Simplify $\displaystyle \int \dfrac{e^{\displaystyle \tan^{-1} \sqrt{x}}}{\sqrt{x} + x\sqrt{x}} \,dx$

$x$ is a literal. The inverse of the tangent of $\sqrt{x}$ is an exponent to the Napier constant and it is divided by the sum of the terms $\sqrt{x}$ and $x\sqrt{x}$. It is required to integrate the function with respect to $x$.

$\displaystyle \int \dfrac{e^{\displaystyle \tan^{-1} \sqrt{x}}}{\sqrt{x} + x\sqrt{x}} \,dx$


Simplify the expression

$\sqrt{x}$ is a common factor in the both terms of the denominator.

$=$ $\displaystyle \int \dfrac{e^{\displaystyle \tan^{-1} \sqrt{x}}}{\sqrt{x}(1+x)} \,dx$


Transformation of the expression

$=$ $\displaystyle \int e^{\displaystyle \tan^{-1} \sqrt{x}} \times \dfrac{1}{\sqrt{x}(1+x)} \,dx$

The part of the expression is the differentiation of the exponent of the Napier constant. So, represent it by any literal.

$u = \tan^{-1} \sqrt{x}$

Differentiate both sides with respect to $x$. It can be differentiated by the differentiation formula of the inverse of tangent function.

$\dfrac{d}{dx} u = \dfrac{d}{dx} \tan^{-1} \sqrt{x}$

$\implies$ $\dfrac{du}{dx} = \dfrac{1}{1+{(\sqrt{x})}^2} \times \dfrac{d}{dx} \sqrt{x}$

$\implies$ $\dfrac{du}{dx} = \dfrac{1}{1+x} \times \dfrac{1}{2\sqrt{x}}$

$\implies$ $\dfrac{du}{dx} = \dfrac{1}{2} \times \dfrac{1}{\sqrt{x}(1+x)}$

$\implies$ $2 du = \dfrac{1}{\sqrt{x}(1+x)} dx$

Now, transform the entire expression in terms of $u$ from $x$.

$=$ $\displaystyle \int e^{\displaystyle u} \times 2 du$

$=$ $\displaystyle 2 \int e^{\displaystyle u} \,du$

Now perform integration.

$=$ $\displaystyle 2(e^{\displaystyle u} +C)$

$=$ $\displaystyle 2e^{\displaystyle u} +2C$

$=$ $\displaystyle 2e^{\displaystyle u} +C$


Required solution

Now, replace the value of $u$ to obtain the required solution for this integration problem.

$=$ $\displaystyle 2e^{\displaystyle \tan^{-1} \sqrt{x}} +C$

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