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Chapter 7: Problem 47

Use integration by parts to derive the following reduction formulas. $$\int \ln ^{n} x d x=x \ln ^{n} x-n \int \ln ^{n-1} x d x$$

Short Answer

Expert verified
Question: Derive the reduction formula for the integral of $$\ln^n x$$ using integration by parts. Solution: The reduction formula for the integral of $$\ln^n x$$ is given by $$\int \ln^{n}x dx = x\ln^{n}x - n \int \ln^{n-1}x dx$$.

Step by step solution

01

Choose u and dv

In order to apply integration by parts, we must select the functions u and dv. Let's choose: $$u = \ln^{n}x$$ $$dv = dx$$
02

Determine du and v

Now we need to find the derivatives and integrals of u and dv, respectively. Compute: $$du = \frac{d (\ln^{n}x)}{dx} = n\ln^{n-1}x \cdot \frac{1}{x}dx$$ $$v = \int dx = x$$
03

Apply Integration by Parts

We have determined the expressions for u, dv, du, and v. Let's substitute these into the integration by parts formula: $$\int \ln^{n}x dx = (\ln^{n}x) \cdot x - \int x \cdot n\ln^{n-1}x \cdot \frac{1}{x} dx$$
04

Simplify the Expression

We can simplify the expression obtained in Step 3: $$\int \ln^{n}x dx = x\ln^{n}x - n \int \ln^{n-1}x dx$$ This is our desired reduction formula.

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Most popular questions from this chapter

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 3 x \cos 7 x d x$$

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