### Video Transcript

Determine the definite integral of

two 𝑥 squared multiplied by 𝑒 to the power of 𝑥 plus two with respect to 𝑥.

In this question we need to find

the integral of a product of functions. As such we will use integration by

parts. This states that for two

differentiable functions 𝑢 and 𝑣, the integral of 𝑢 multiplied by d𝑣 by d𝑥 with

respect to 𝑥 is equal to 𝑢𝑣 minus the integral of 𝑣 multiplied by d𝑢 by d𝑥

with respect to 𝑥. This is also written as the

integral of 𝑢𝑣 prime is equal to 𝑢𝑣 minus the integral of 𝑣𝑢 prime. We begin this process by choosing

the functions 𝑢 and d𝑣 by d𝑥.

The LIATE rule tells us to choose

𝑢 to be the function that appears first in the list: logarithmic functions, inverse

trigonometric functions, algebraic functions, trigonometric functions, exponential

functions. Our integrand is the product of a

polynomial or algebraic function and an exponential function. Since A occurs before E, we choose

the algebraic function two 𝑥 squared to be 𝑢. Hence, we set d𝑣 by d𝑥 equal to

𝑒 to the power of 𝑥 plus two. Next, we find d𝑢 by d𝑥 by

differentiating 𝑢 and 𝑣 by integrating d𝑣 by d𝑥. Using the power rule of

differentiation, we can calculate the derivative of two 𝑥 squared, which is four

𝑥.

Next, recalling the general rule

for integrating exponential functions, where the integral of 𝑒 to the power of 𝑥

is equal to 𝑒 to the power of 𝑥. And also recalling the laws of

exponents, we see that the integral of 𝑒 to the power of 𝑥 plus two is equal to 𝑒

squared multiplied by 𝑒 to the power of 𝑥, which can be rewritten as 𝑒 to the

power of 𝑥 plus two, and this is our value of 𝑣. We are now in a position to

substitute our expressions into the formula for integration by parts.

The integral of two 𝑥 squared

multiplied by 𝑒 to the power of 𝑥 plus two with respect to 𝑥 is equal to two 𝑥

squared multiplied by 𝑒 to the power of 𝑥 plus two minus the integral of four 𝑥

multiplied by 𝑒 to the power of 𝑥 plus two with respect to 𝑥. Since the new integrand is still

the product of functions, we need to repeat the process and use integration by parts

once again. This time, 𝑢 is equal to four 𝑥

and d𝑣 by d𝑥 is equal to 𝑒 to the power of 𝑥 plus two.

Differentiating four 𝑥, we have

d𝑢 by d𝑥 is equal to four. And integrating 𝑒 to the power of

𝑥 plus two, we have 𝑣 is equal to 𝑒 to the power of 𝑥 plus two. Substituting in these values, we

have two 𝑥 squared multiplied by 𝑒 to the power of 𝑥 plus two minus four 𝑥

multiplied by 𝑒 to the power of 𝑥 plus two minus the integral of four 𝑒 to the

power of 𝑥 plus two with respect to 𝑥. This can be simplified as

shown.

Next, we need to integrate four 𝑒

to the power of 𝑥 plus two with respect to 𝑥. We now have two 𝑥 squared

multiplied by 𝑒 to the power of 𝑥 plus two minus four 𝑥 multiplied by 𝑒 to the

power of 𝑥 plus two plus four multiplied by 𝑒 to the power of 𝑥 plus two. Next, we will factor out the common

factor of 𝑒 to the power of 𝑥 plus two. And we will also factor out the

constant four, giving us four multiplied by a half 𝑥 squared minus 𝑥 plus one

multiplied by 𝑒 to the power of 𝑥 plus two.

Recalling that we need to add our

constant of integration 𝐶, we have the final answer to the integral of two 𝑥

squared multiplied by 𝑒 to the power of 𝑥 plus two with respect to 𝑥. It is four multiplied by a half 𝑥

squared minus 𝑥 plus one multiplied by 𝑒 to the power of 𝑥 plus two plus 𝐶.