We want to compute the integral int (5x^(2)+2x+3)e^(5x)dx using two (and only two) integration by parts. This means that you have to make a good
choice of f and g^(') for your two integration by parts.
For the first integration by parts, you should choose
[f(x),g^(')(x)]=
Warning: for this question, you must use strict calculator notations and also include the square brackets in your answer; in particular, you must use * for
every multiplication (e.g. 2x is written 2*x.)
to get
int (5x^(2)+2x+3)e^(5x)dx=G(x)-int H(x)dx
where
G(x)=
and
H(x)=
Now, to integrate int H(x)dx, we need to use the method of integration by parts a second time with
[f(x),g^(')(x)]=
Warning: for this question, you must use strict calculator notations and also include the square brackets in your answer; in particular, you must use * for
every multiplication (e.g. 2x is written 2*x.)
to get
int H(x)dx=, 욜 +C
(Don't add the constant of integration C since we have done it for you.)
We have then found that
int (5x^(2)+2x+3)e^(5x)*dx=
choice of f and g for your two integration by parts. For the first integration by parts, you should choose [fgx]= Warning: for this question, you must use strict calculator notations and also include the square brackets in your answer in particular, you must use * for every multiplication e.g.2 is written 2*x. to get 52+2x+3e5x=Gx-[Hdx where Gx= and Hx=
Now, to integrate )d,we need to use the method of integration by parts a second time with
[fxgx]= 40 Warning:for this question, you must use strict calculator notations and also include the square brackets in your answer in particular, you must use *for every multiplication e.g.2 is written 2*x.) A 1060 Hd= BE+C (Don't add the constant of integration C since we have done it for you.) We have then found that 5+2x+3ebx= BB+C