Question

How do you evaluate $\int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} $?

Answer 1

Hint: We can use Integration by substitution to solve this integral. So, first substitute $u = \sqrt x $ and find its differentiation with respect to $x$ using differentiation properties. Then substitute the value of $x$ and value of $dx$ in the given integral. Next, integrate using integration properties. Finally, substitute the value of $u$, and get the desired result.

Formula used: The differentiation of the product of a constant and a function = the constant $ \times $ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
Differentiation formula: $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$
The integral of the product of a constant and a function = the constant $ \times $ integral of the function.
i.e., $\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx} $, where $k$ is a constant.
The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx} $
Integration formula: $\int {{{\tan }^{ - 1}}xdx} = x{\tan ^{ - 1}}x - \dfrac{1}{2}\ln \left( {1 + {x^2}} \right) + C$

Complete step-by-step solution:
We have to find $\int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} $...............…(i)
We will use Integration by substitution to solve this integral.
So, for the integral involving the root$\sqrt x $, we use the substitution:
$u = \sqrt x $...............…(ii)
Now, we have to differentiate $u$ with respect to $x$.
So, differentiating $u$ with respect to $x$, we get
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {\sqrt x } \right)$...............…(iii)
Now, using the differentiation formula $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$ in above differentiation and find the value of $dx$.
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{2\sqrt x }}$
$ \Rightarrow dx = 2\sqrt x du$...............…(iv)
Now, we have to substitute the value of $\sqrt x $ from (ii) and the value of $dx$ from (iv) in integral (i).
\[\int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = \int {\dfrac{{\arctan \left( u \right)}}{{\sqrt x }} \times 2\sqrt x du} \]
$ \Rightarrow \int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = \int {\arctan \left( u \right) \times 2du} $...............…(v)
Now, using the property that the integral of the product of a constant and a function = the constant $ \times $ integral of the function.
i.e., $\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx} $, where $k$ is a constant.
So, in above integral (v), constant $2$ can be taken outside the integral.
$ \Rightarrow \int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2\int {\arctan \left( u \right)du} $..............…(vi)
Now, using the integration formula $\int {{{\tan }^{ - 1}}xdx} = x{\tan ^{ - 1}}x - \dfrac{1}{2}\ln \left( {1 + {x^2}} \right) + C$ in integral (vi), we get
$ \Rightarrow \int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2\left[ {u\arctan \left( u \right) - \dfrac{1}{2}\ln \left( {1 + {u^2}} \right)} \right] + C$
$ \Rightarrow \int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2u\arctan \left( u \right) - \ln \left( {1 + {u^2}} \right) + C$…(vii)
We have to find the integral in terms of $x$. So, replacing $u$ with $x$ using (ii).
$ \Rightarrow \int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2\sqrt x \arctan \left( {\sqrt x } \right) - \ln \left| {x + 1} \right| + C$

Hence, $\int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2\sqrt x \arctan \left( {\sqrt x } \right) - \ln \left| {x + 1} \right| + C$.

Note: We can also solve the given integral by Integration by parts: $\int {udv} = uv - \int {vdu} $.
Use integration by parts with $u = \arctan \left( {\sqrt x } \right)$ and $dv = \dfrac{1}{{\sqrt x }}dx$
Differentiate $u$ with respect to $x$.
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {\arctan \sqrt x } \right)$
Now, using the differentiation formula $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$ in above differentiation, we get
$\dfrac{{du}}{{dx}} = \dfrac{1}{{1 + x}} \times \dfrac{1}{{2\sqrt x }}$
$ \Rightarrow du = \dfrac{1}{{2\sqrt x \left( {x + 1} \right)}}dx$
Now, integrate $v$ with respect to $x$.
$\int {dv} = \int {\dfrac{1}{{\sqrt x }}dx} $
Now, using the integration formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1$ in above integral, we get
$v = 2\sqrt x $
The integration by parts formula is:
$\int {udv} = uv - \int {vdu} $
Put the value of $u,v,du,dv$.
$\int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2\sqrt x \arctan \left( {\sqrt x } \right) - \int {2\sqrt x \times \dfrac{1}{{2\sqrt x \left( {x + 1} \right)}}dx} $
$ \Rightarrow \int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2\sqrt x \arctan \left( {\sqrt x } \right) - \int {\dfrac{1}{{\left( {x + 1} \right)}}dx} $
Now, using the integration formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1$ in above integral, we get
$ \Rightarrow \int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2\sqrt x \arctan \left( {\sqrt x } \right) - \ln \left| {x + 1} \right| + C$
Hence, $\int {\dfrac{{\arctan \left( {\sqrt x } \right)}}{{\sqrt x }}dx} = 2\sqrt x \arctan \left( {\sqrt x } \right) - \ln \left| {x + 1} \right| + C$.