Question

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

Answer 1

Hint: 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:
\[ \Rightarrow \dfrac{1}{a}\int {\dfrac{1}{{1 + {u^2}}}} du = \dfrac{1}{a}\arctan u = \dfrac{1}{a}\arctan \dfrac{x}{a} + C\]