Question
Question: How do you integrate \(\int {{x^2}{e^{2x}}} \) by parts?...
How do you integrate ∫x2e2x by parts?
Solution
This question will be solved by integration by parts. In the integration by parts method if we integrate f(x)g(x) we can write ∫f(x)g(x)dx=f(x)∫g(x)dx−∫f′(x)(∫g(x)dx)dx where f′(x)=dxdf(x). While choosing f(x) and g(x) choose in such a way that solving ∫f′(x)(∫g(x)dx)dx would be easier. For example, while integrating xex our f(x) would be x and g(x) will be ex so that solving ∫f′(x)(∫g(x)dx)dx is easier.
Complete step-by-step answer:
The integration by parts formula can be further written as integral of the product of any two functions = (First function × Integral of the second function) – Integral of [ (differentiation of the first function) × Integral of the second function]
The formula for integration by parts is
∫f(x)g(x)dx=f(x)∫g(x)dx−∫f′(x)(∫g(x)dx)dx
Where f′(x)=dxdf(x).
In the given question we have to integrate x2e2x by integration by parts method.
So, we can choose f(x)=x2 and g(x)=e2x.
Substituting the values in the formula for integration by parts, we get
⇒∫x2e2xdx=x2∫e2xdx−∫dxd(x2)(∫e2xdx)dx
We know that dxd(x2)=2x and ∫e2xdx=2e2x. Substituting these values in the above equation,
⇒∫x2e2xdx=2x2e2x−∫2x×2e2xdx
Simplify the terms,
⇒∫x2e2xdx=2x2e2x−∫xe2xdx ….. (1)
Now apply integration by parts for xe2x.
⇒∫xe2xdx=x∫e2xdx−∫dxdx(∫e2xdx)dx
Simplify the terms,
⇒∫xe2xdx=2xe2x−∫2e2xdx
Integrate the term,
⇒∫xe2xdx=2xe2x−4e2x
Substitute the value in equation (1),
⇒∫x2e2xdx=2x2e2x−2xe2x+4e2x+C
Hence, the integral of x2e2x is ∫x2e2xdx=2x2e2x−2xe2x+4e2x+C.
Note:
Always remember the formula for integrating into the integration by parts method. Some people make mistakes while choosing f(x) and g(x) so carefully choose f(x) and g(x) such that integration would be easier to solve. Sometimes when we solve a problem we find the function of L.H.S on the right-hand side, in that case, we should take our L.H.S as a variable I like we did in the above question then it would be easier to solve we just have to find the value of I.