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Chapter 6: Problem 17

Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int\left(x^{2}-1\right) e^{x} d x $$

Short Answer

Expert verified
The integral \(\int (x^2 - 1) e^x dx\) is equal to \(x^2 e^x - 2x e^x + 2e^x + C\)

Step by step solution

01

Use the integration by parts formula

The formula for integration by parts is \(\int u dv = uv - \int v du\). Here, set \(u = x^2 - 1\), which differentiates easily and \(dv = e^x dx\), which integrates easily. So, \(du = 2x dx\) and \(v = e^x\).
02

Apply the formula

Insert \(u\), \(v\), \(du\), and \(dv\) into the integration by parts formula. This results in: \(\int (x^2 - 1) e^x dx = (x^2 - 1)e^x - \int e^x\cdot2x dx\).
03

Notice the integral in Step 2 requires integration by parts again

The integral in the formula from Step 2 again contains a product of an exponential function and a linear function. Therefore, Integration by parts will be applied again. For the second integration by parts, choose \(u = 2x\) and \(dv = e^x dx\), then \(du = 2 dx\) and \(v = e^x\).
04

Apply the integration by parts formula again

Insert new \(u\), \(v\), \(du\) and \(dv\) into integration by parts formula, you will get : \(\int 2x e^x dx = 2x e^x - 2 \int e^x dx = 2x e^x - 2e^x + C\) where C is the constant of integration.
05

Substitute this back into the equation from Step 2

We now have: \((x^2 - 1)e^x - \int 2x e^x dx = (x^2 - 1)e^x - (2x e^x - 2e^x + C)\). Simplify to get the final answer.

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Most popular questions from this chapter

(a) Use a graphing utility to graph the function \(y=e^{-x^{2}}\). (b) Show that \(\int_{0}^{\infty} e^{-x^{2}} d x=\int_{0}^{1} \sqrt{-\ln y} d y\).

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