# Question Video: Finding the Integration of a Function Involving an Exponential Function Using Integration by Parts Twice

Integrating Exponential Functions (Basics)
Integrating Exponential Functions (Basics)

### 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.

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