We want to compute the integral ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(4x^2+3x+7)e^{4x}dx$ using two $($and only two$)$ integration by parts. This means that you have to make a good choice of $f$ and $g^{\text{'}}$ for your two integration by parts.

For the first intagration by parts, you should choose

$[f(x),g^{\text{'}}(x)]=$

$$

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 multipliaction $($e$.$g$.2x$ is written $2^{**}x.)$ to get

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(4x^2+3x+7)e^{4x}dx=G(x)-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}H(x)dx$

where

$G(x)=$

$$

and

$H(x)=$

$$

Now, to integrate ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}H(x)dx,$ we need to use the method of integration by parts a second time with

$[f(x),g^{\text{'}}(x)]=$

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 multipliaction $($e$.$g$.2x$ is written $2^{**}x.)$

to get

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}H(x)dx=$

$+C$

$($Don$\text{'}$t add the constant of integration $C$ since we have done it for you.$)$

We have then found that

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(4x^2+3x+7)e^{4x}dx=$

$+C$