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Q. 51

Expert-verified
Found in: Page 362

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess- and- check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. (Hint for Exercise 54: $\mathrm{tan}\left(x\right)=\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}$). $\int {x}^{2}{\left({x}^{3}+1\right)}^{5}dx.$

The value of given integral is $\frac{{\left({x}^{3}+1\right)}^{6}}{18}+c.$

See the step by step solution

## Step 1. Given Information.

Given an integral$\int {x}^{2}{\left({x}^{3}+1\right)}^{5}dx.$

## Step 2. Formula involved.

We use substitution method in this integration and then below formula will be used:

$\int {x}^{n}dx=\frac{{x}^{n}}{n+1}+c.$

## Step 3. Solving the integral.

$Lett={x}^{3}+1,\phantom{\rule{0ex}{0ex}}dt=3{x}^{2}dx,or{x}^{2}dx=\frac{dt}{3}.\phantom{\rule{0ex}{0ex}}Subsitutingthisintheintegralweget,\phantom{\rule{0ex}{0ex}}\int {x}^{2}{\left({x}^{3}+1\right)}^{5}dx\phantom{\rule{0ex}{0ex}}=\int {t}^{5}\frac{dt}{3}=\frac{1}{3}\int {t}^{5}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\frac{{t}^{6}}{6}+c=\frac{{t}^{6}}{18}+c=\frac{{\left({x}^{3}+1\right)}^{6}}{18}+c.$