Solveeit Logo

Question

Question: How do you integrate by parts: \[\int {x{e^3}x\,dx} \]?...

How do you integrate by parts: xe3xdx\int {x{e^3}x\,dx} ?

Explanation

Solution

Here in this question given an indefinite integral, we have to find the integrated value of given function. It can be solved by the method of integration by parts by separating the function as uu and vv, later integrated by using the standard formulas of integration. And by further simplification we get the required solution.

Complete step by step solution:
Consider the given integral function:
xe3xdx\Rightarrow \,\,\,\int {x{e^3}x\,dx}---------(1)
Here, e3{e^3} is a constant term then take it outside from the integral, then equation (1) becomes
e3x2dx\Rightarrow \,\,\,{e^3}\int {{x^2}dx}-----(2)
On integrating this directly we get
e3(x33)+c\Rightarrow \,\,\,{e^3}\left( {\dfrac{{{x^3}}}{3}} \right) + c
But in the question, they have mentioned we have to solve the given integral using the method of integration by parts.
Integration by parts is a technique for performing indefinite integration or definite integration by expanding the differential of a product of functions d(uv)d\left( {uv} \right) and expressing the original integral in terms of a known integral vdu\int {v\,du} . A single integration by parts starts with
d(uv)=udv+vdud(uv) = u\,dv + v\,du
and integrates both sides,
d(uv)=uv=udv+vdu.\int {d(uv)} = uv = \int {u\,dv} + \int {v\,} du.
Rearranging gives
udv=uvvdu.\int {udv = uv - \int {vdu.} } ------(3)
Apply the equation (3) in equation (2), then
Let take, u=x2dudx=2xdu=2xdxu = {x^2} \Rightarrow \,\dfrac{{du}}{{dx}} = 2x \Rightarrow du = 2x\,dx
and dvdx=1v=x\dfrac{{dv}}{{dx}} = 1 \Rightarrow v = x
e3x2dx=e3(x2xx2xdx)\Rightarrow \,\,\,{e^3}\int {{x^2}dx} = \,{e^3}\left( {\,{x^2} \cdot x - \int {x\, \cdot } 2x\,dx} \right)
Where e3{e^3} is constant term
e3x2dx=e3(x32x2dx)\Rightarrow \,\,\,{e^3}\int {{x^2}dx} = \,{e^3}\left( {\,{x^3} - 2\int {{x^2}\,} \,dx} \right)
e3x2dx=e3(x32x33)+c\Rightarrow \,\,\,{e^3}\int {{x^2}dx} = \,{e^3}\left( {\,{x^3} - 2\dfrac{{{x^3}}}{3}\,} \right) + c
On simplification, we get
e3x2dx=e3(3x32x33)+c\Rightarrow \,\,\,{e^3}\int {{x^2}dx} = \,{e^3}\left( {\,\dfrac{{3{x^3} - 2{x^3}}}{3}\,} \right) + c
e3x2dx=e3(x33)+c\Rightarrow \,\,\,{e^3}\int {{x^2}dx} = \,{e^3}\left( {\,\dfrac{{{x^3}}}{3}\,} \right) + c
Where C is an integrating constant.
Hence, the value of xe3xdx\int {x{e^3}x\,dx} is e3(x33)+c\,{e^3}\left( {\,\dfrac{{{x^3}}}{3}\,} \right) + c.

Note: In integration we have two kinds one is definite integral and other one is indefinite integral. This question comes under the indefinite integral. While integrating the function which is in the form of product or division form we use the integration by parts method. By applying the integration by parts we obtain the solution.