Question

How do you use substitution to integrate $x\sqrt {2x + 1} dx$?

Answer 1

We have been given an integration $\int {f\left( x \right)dx}$ . To evaluate the integral using integration by substitution method, we transformed the given integral into another form by changing the independent variable $x$ into $t$ by using $x = g\left( t \right)$ such that after the substitution, the given integral is converted into directly known integrals.

Complete step-by-step solution:
Step 1: Given integral is $\int {x\sqrt {2x + 1} dx}$. Let us consider that given integration is equal to $I$, so we get
$I = \int {x\sqrt {2x + 1} dx}$
Now, we substitute $2x + 1 = t$ .
Now to replace the value of $dx$, we differentiate the $2x + 1 = t$ with respect to $x$ , we get
$2dx = dt \\\ dx = \dfrac{{dt}}{2} \\\$
and to replace the value of $x$ , simplifying the equation , we get
$\Rightarrow 2x = t - 1 \\\ \Rightarrow x = \dfrac{{t - 1}}{2} \\\$
Step 2: Now substituting all the values in the given integral, we get
$I = \int {x\sqrt {2x + 1} dx} \\\ = \int {\left( {\dfrac{{t - 1}}{2}} \right)} \times \sqrt t \times \dfrac{{dt}}{2} \\\$
Step 3: Now multiplying the brackets, we get $I = \dfrac{1}{4}\int {\left( {{t^{1 + \dfrac{1}{2}}} - {t^{\dfrac{1}{2}}}} \right)} dt$
Adding the powers, we get
\Rightarrow$$$I = \dfrac{1}{4}\int {\left( {{t^{\dfrac{3}{2}}} - {t^{\dfrac{1}{2}}}} \right)} dt$$ Step 4: Now integrating the above expression using the standard formula of integration, we get \RightarrowI = \dfrac{1}{4}\left( {\dfrac{{{t^{\dfrac{3}{2} + 1}}}}{{\dfrac{3}{2} + 1}} - \dfrac{{{t^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}}} \right) + C$$ Adding the powers in the numerator and adding the numbers in denominator, we get $\RightarrowI = \dfrac{1}{4}\left( {\dfrac{{{t^{\dfrac{5}{2}}}}}{{\dfrac{5}{2}}} - \dfrac{{{t^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right) + C$$
Now simplifying the above expression, we get

$$\Rightarrow I = \dfrac{1}{4}\left( {\dfrac{{2{t^{\dfrac{5}{2}}}}}{5} - \dfrac{{2{t^{\dfrac{3}{2}}}}}{3}} \right) + C \\\ \Rightarrow I = \dfrac{1}{2}\left( {\dfrac{{{t^{\dfrac{5}{2}}}}}{5} - \dfrac{{{t^{\dfrac{3}{2}}}}}{3}} \right) + C \\\$$

Step 5: Now to get the original integrand, we replace the value of $t$ .
In the beginning we have consider the value of $t = 2x + 1$ , so replacing the value of $t$ , we get
$\Rightarrow I = \dfrac{1}{2}\left( {\dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{5}{2}}}}}{5} - \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{3}} \right) + C$
Taking ${\left( {2x + 1} \right)^{\dfrac{3}{2}}}$ common from numerator, we get
$\Rightarrow I = \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{2}\left( {\dfrac{{\left( {2x + 1} \right)}}{5} - \dfrac{1}{3}} \right) + C$
Step 6: Taking LCM inside the bracket and simplify, we get

$$\Rightarrow I = \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{2}\left( {\dfrac{{3\left( {2x + 1} \right) - 5}}{{15}}} \right) + C \\\ \Rightarrow I = \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{2}\left( {\dfrac{{6x + 3 - 5}}{{15}}} \right) + C \\\ \Rightarrow I = \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{2}\left( {\dfrac{{6x - 2}}{{15}}} \right) + C \\\$$

Now take $2$ common from the bracket, we get

$$\Rightarrow I = \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{2}\left( {\dfrac{{2\left( {3x - 1} \right)}}{{15}}} \right) + C \\\ \Rightarrow I = {\left( {2x + 1} \right)^{\dfrac{3}{2}}}\left( {\dfrac{{3x - 1}}{{15}}} \right) + C \\\$$

This is the final integration of the given function.

Note: Formula to integrate ${x^n}$ is given as
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$ , where $C$ is the constant of integration.
It is important to guess the useful substitution function. Choose the substitution function such that its derivative also occurs in the integrand.