Question

Integrate $\left( {\dfrac{{\sqrt {x + 1} }}{x}} \right)dx$.

Answer 1

Hint: As, the function is not continuous. Substitute the numerator to any variable to get rid of the square root. After that transform the denominator and $dx$ in terms of the variable. Then, add and subtract a constant to simplify the terms. After that, split the remaining fraction. Then integrate the term to get the desired result.

Complete step by step answer:
It is given in the question that we have to integrate $\left( {\dfrac{{\sqrt {x + 1} }}{x}} \right)dx$.
Since, the function is not continuous. Substitute the numerator to any variable to get rid of the square root,
$ \Rightarrow \sqrt {x + 1} = u$
Square the terms on both sides,
$ \Rightarrow x + 1 = {u^2}$
Move 1 to the right side,
$ \Rightarrow x = {u^2} - 1$
Now differentiate on both sides,
$ \Rightarrow dx = 2udu$
Substitute these values in the integration,
$ \Rightarrow \int {\dfrac{{\sqrt {x + 1} }}{x}dx} = \int {\dfrac{u}{{{u^2} - 1}} \times 2udu} $
Simplify the terms on the right side,
$ \Rightarrow \int {\dfrac{{\sqrt {x + 1} }}{x}dx} = 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} $
Since, both numerator and denominator have the same degree. Add and subtract 1 in the numerator,
$ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2\int {\dfrac{{{u^2} - 1 + 1}}{{{u^2} - 1}}du} $
Now, split the terms,
$ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2\int {\dfrac{{{u^2} - 1}}{{{u^2} - 1}}du} + 2\int {\dfrac{1}{{{u^2} - 1}}du} $
Cancel out common factors,
$ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2\int {1du} + 2\int {\dfrac{1}{{{u^2} - 1}}du} $
Now integrate first part and apply the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ in the denominator of the second part,
$ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2u + \int {\dfrac{2}{{\left( {u - 1} \right)\left( {u + 1} \right)}}du} $
Now, split the fraction,
$ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2u + \int {\left( {\dfrac{1}{{u - 1}} - \dfrac{1}{{u + 1}}} \right)du} $
Again, split the function in two parts,
$ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2u + \int {\dfrac{1}{{u - 1}}du} - \int {\dfrac{1}{{u + 1}}du} $
Now integrate the terms,
$ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2u + \ln \left| {u - 1} \right| - \ln \left| {u + 1} \right| + C$
Apply the log formula, $\log A - \log B = \log \dfrac{A}{B}$
$ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2u + \ln \dfrac{{\left| {u - 1} \right|}}{{\left| {u + 1} \right|}} + C$
Substitute back the value of $u$,
\[ \Rightarrow 2\int {\dfrac{{{u^2}}}{{{u^2} - 1}}du} = 2\sqrt {x + 1} + \ln \dfrac{{\left| {\sqrt {x + 1} - 1} \right|}}{{\left| {\sqrt {x + 1} + 1} \right|}} + C\]

Hence, the integration of $\left( {\dfrac{{\sqrt {x + 1} }}{x}} \right)dx$ is \[2\sqrt {x + 1} + \ln \dfrac{{\left| {\sqrt {x + 1} - 1} \right|}}{{\left| {\sqrt {x + 1} + 1} \right|}} + C\].

Note: Integration is often wont to find areas, volumes, central points, and lots of useful things. it's often wont to find the world under the graph of a function. The symbol for "Integral" may be a stylish "S" (for "Sum", the thought of summing slices).
This question is often solved in only a couple of steps by using basic integration rules and formulas. So, confirm to memorize all the essential integration formulae. This may save time for solving the question. Also, integration is the reverse of differentiation. Check if the obtained answer is correct by differentiating the obtained result with reference to $x$.