Question

How do you find the integral of $\int {\dfrac{1}{{\sqrt x \times \left( {1 + x} \right)}}} dx$ ?

Answer 1

From the given function first declare a variable $u$ and substitute it into the integral then Differentiate $u$ and isolate the $x$ term. This gives you the differential $du = dx$ . Substitute $du$ for $dx$ in the integral: Evaluate the integral and Substitute back $x$ value in the place of $u$

Complete step-by-step solution:
To integrate the given equation
$\Rightarrow \int {\dfrac{1}{{\sqrt x \times \left( {1 + x} \right)}}} dx$
Consider $\sqrt x = u$
We can write $u$ as
$\Rightarrow u = {\left( x \right)^{\dfrac{1}{2}}}$
Therefore on differentiating $u$ , we get
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{2}{\left( x \right)^{\dfrac{1}{2} - 1}}$
Now solve power value
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{2}{\left( x \right)^{ - \dfrac{1}{2}}}$
So therefore we can write the component as
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{2\sqrt x }}$
Now bring $dx$ to the RHS, we get
$\Rightarrow du = \dfrac{1}{{2\sqrt x }}dx$
Now substitute $u$ in the place of $\sqrt x$
Therefore we get,
$\Rightarrow du = \dfrac{1}{{2u}}dx$
Now find the value of $dx$ we get
$\Rightarrow dx = 2udu$
Substitute this $u$ and $dx$ value in the given equation
Therefore we get
$\Rightarrow \int {\dfrac{1}{{u \times \left( {1 + {u^2}} \right)}}} 2u\,du$
Now cancel out $u$
$\Rightarrow 2\int {\dfrac{1}{{1 + {u^2}}}} du$
Substitute the value of $\int {\dfrac{1}{{1 + {u^2}}}} du$
Then we get
$\Rightarrow 2\,\arctan \left( u \right)$
Now substitute the value of $u$
Therefore we get
$\Rightarrow 2\,\arctan \sqrt x + C$

Hence the integral of $\int {\dfrac{1}{{\sqrt x \times \left( {1 + x} \right)}}} dx$ is $2\,\arctan \sqrt x + C$

Note: The following integral is very common in calculus:
$\Rightarrow \int {\dfrac{1}{{1 + {x^2}}}} dx = \arctan x + C$
A more general form is
$\Rightarrow \int {\dfrac{1}{{{a^2} + {x^2}}}} dx = \dfrac{1}{a}\arctan \left( {\dfrac{x}{a}} \right) + C$
Proof:
Factor ${a^2}$ from the denominator:
$\Rightarrow \int {\dfrac{1}{{{a^2} + {x^2}}}} dx = \int {\dfrac{1}{{{a^2}\left( {1 + \dfrac{{{x^2}}}{{{a^2}}}} \right)}}dx = } \dfrac{1}{{{a^2}}}\int {\dfrac{1}{{1 + \left( {\dfrac{{{x^2}}}{{{a^2}}}} \right)}}dx}$
Now we do a $udu$ substitution, with $u = \dfrac{x}{a}$ so that $du = \dfrac{1}{a}dx$
Thus, $dx = adu$
We make the replacements:
$\Rightarrow \dfrac{1}{{{a^2}}}\int {\dfrac{1}{{\left( {1 + \left( {\dfrac{{{x^2}}}{{{a^2}}}} \right)} \right)}}} dx = \dfrac{1}{{{a^2}}}\int {\dfrac{1}{{1 + {u^2}}}} \left( {a\,du} \right)$
Note that the $a$ inside the integral comes out to the front, so we have
$\Rightarrow \dfrac{1}{{{a^2}}}\int {\dfrac{1}{{1 + {u^2}}}} \left( {a\,du} \right) = \dfrac{1}{a}\int {\dfrac{1}{{1 + {u^2}}}} du$
Now we integrate:
$\Rightarrow \dfrac{1}{a}\int {\dfrac{1}{{1 + {u^2}}}} du = \dfrac{1}{a}\arctan u = \dfrac{1}{a}\arctan \dfrac{x}{a} + C$