Question

How do you find the integral of (2x^3+5x+1)e^(2x) dx ?

Answer 1

expand the equation first
[(2x^3)e^(2x)+5x*e^(2x)+e^(2x)] dx

\[\rightarrow \int\limits_{}^{}(2x^3)e ^{2x} dx +\int\limits_{}^{}5x \times e ^{2x} dx + \int\limits_{}^{} e ^{2x}dx\]

to the integration by part seperately

Answer 2

integration by parts three times on the first integral, integrate by parts once on the second integral, and use u substitution on the third integral, then add them all together

Answer 3

first integral, Integrate by parts 3 times:
\[\int\limits\limits 2x^3e^{2x}dx \rightarrow\]
\[f(x)=e^{2x}, F(x)=\frac{e^{2x}}{2}, g(x)=x^3, g'(x)=3x^2\]
\[2 ( \frac{x^3e^{2x}}{2}- \frac{3}{2}\int\limits\limits\limits x^2e^{2x}dx)\rightarrow \]\[x^3e^{2x}-3 \int\limits\limits x^2e^{2x}dx\]
\[f(x)=e^{2x}, F(x)=\frac{e^{2x}}{2}, g(x)=x^2, g'(x)=2x\]
\[x^3e^{2x}-3( \frac{x^2e^{2x}}{2}- \int\limits\limits\limits xe^{2x}dx)\rightarrow \]\[x^3e^{2x}-\frac{3}{2}x^2e^{2x}+3 \int\limits\limits xe^{2x}dx\]
\[f(x)=e^{2x}, F(x)=\frac{e^{2x}}{2}, g(x)= x, g'(x)=1\]
\[x^3e^{2x}-\frac{3}{2}x^2e^{2x}+3( \frac{xe^{2x}}{2}-\frac{1}{2} \int\limits\limits\limits e^{2x}dx) \rightarrow\]
\[x^3e^{2x}-\frac{3}{2}x^2e^{2x}+\frac{3}{2}xe^{2x}-\frac{3}{2}e^{2x}+C\]
second integral, integrate by parts once:
\[\int\limits 5xe^{2x}dx \rightarrow\]
\[f(x)=e^{2x}, F(x)=\frac{e^{2x}}{2}, g(x)=x, g'(x)=1\]
\[5( \frac{xe^{2x}}{2}- \frac{1}{2} \int\limits e^{2x}dx)\rightarrow \frac{5}{2}xe^{2x}- \frac{5}{2}e^{2x}+C\]
third integral, use u substitution:
\[\int\limits e^{2x}dx \rightarrow \frac{1}{2}e^{2x} +C\]
Your final answer should be:
\[x^3e^{2x}-\frac{3}{2}x^2e^{2x}+\frac{3}{2}xe^{2x}-\frac{3}{2}e^{2x}+\frac{5}{2}xe^{2x}\]
\[- \frac{5}{2}e^{2x}+\frac{1}{2}e^{2x}+C\]

Answer 4

The final answer can be simplified, which I'll leave up to you

Answer 5

THANKS!!

Answer 6

no problem