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Question: How do you use substitution to integrate \(x\left( {\sqrt {2x + 1} } \right)dx\)?...

How do you use substitution to integrate x(2x+1)dxx\left( {\sqrt {2x + 1} } \right)dx?

Explanation

Solution

We will first substitute the expression inside the square root to be something and then find x according to that. Now, we will find dx and put all of them so that we get the required answer.

Complete step-by-step solution:
We are given that we are required to solve the integration of x(2x+1)dxx\left( {\sqrt {2x + 1} } \right)dx using the substitution method. This means that we need to find the value of I=∫x(2x+1)dxI = \int {x\left( {\sqrt {2x + 1} } \right)dx} ……………..(1)
We will first substitute 2x + 1 to be t2{t^2}.
Now, since we have 2x+1=t22x + 1 = {t^2} ……………(2)
Taking the 1 from addition in the left hand side of the above equation to subtraction in the right hand side, we will then obtain the following expression:-
⇒2x=t2−1\Rightarrow 2x = {t^2} - 1
Now, if we divide both the sides of the above equation by 2, we will then obtain the following expression:-
⇒x=t2−12\Rightarrow x = \dfrac{{{t^2} - 1}}{2} ………………..(3)
Differentiating both the sides of the above equation, we will then obtain the following expression:-
⇒ddx(x)=ddt(t22−12)\Rightarrow \dfrac{d}{{dx}}\left( x \right) = \dfrac{d}{{dt}}\left( {\dfrac{{{t^2}}}{2} - \dfrac{1}{2}} \right)
⇒dx=tdt\Rightarrow dx = tdt …………..(4)
Putting the equation number 2, 3 and 4 in equation number 1, we will then obtain the following equation:-
⇒I=∫x(2x+1)dx=∫(t2−12×t×t)dt\Rightarrow I = \int {x\left( {\sqrt {2x + 1} } \right)dx} = \int {\left( {\dfrac{{{t^2} - 1}}{2} \times t \times t} \right)dt}
We can write the above integral as following:-
⇒I=∫t42dt−∫t22dt\Rightarrow I = \int {\dfrac{{{t^4}}}{2}dt} - \int {\dfrac{{{t^2}}}{2}dt}
We can write the above integral as following:-
⇒I=12[∫t4dt−∫t2dt]\Rightarrow I = \dfrac{1}{2}\left[ {\int {{t^4}dt} - \int {{t^2}dt} } \right]
Now, we will use the fact that: ∫xndx=xn+1n+1+C\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C
⇒I=12[t55−t33]+C\Rightarrow I = \dfrac{1}{2}\left[ {\dfrac{{{t^5}}}{5} - \dfrac{{{t^3}}}{3}} \right] + C
Simplifying the brackets, we will then obtain the following equation:-
⇒I=t510−t36+C\Rightarrow I = \dfrac{{{t^5}}}{{10}} - \dfrac{{{t^3}}}{6} + C
Putting back the value of t from equation number 2, we will then obtain the following equation:-
⇒I=(2x+1)5210−(2x+1)326+C\Rightarrow I = \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{5}{2}}}}}{{10}} - \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{6} + C

Thus, (2x+1)5210−(2x+1)326+C\dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{5}{2}}}}}{{10}} - \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{6} + C is the required answer.

Note: The students must note that we used some facts in between like the fact that:-
∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx\int {\left[ {f(x) + g(x)} \right]} dx = \int {f(x)dx} + \int {g(x)dx}
It explains that the integration of the sum of two functions is equal to the sum of integration of those two functions.
ddx[f(x)+g(x)]=ddxf(x)+ddxg(x)\dfrac{d}{{dx}}[f(x) + g(x)] = \dfrac{d}{{dx}}f(x) + \dfrac{d}{{dx}}g(x)
It explains that the differentiation of the sum of two functions is equal to the sum of differentiation of those two functions.
The students must also commit to memory the following formula:-
∫xndx=xn+1n+1+C\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C