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Found in: Page 306

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the integral by making the given substitution.$$\int {{{\rm{x}}^{\rm{2}}}} \sqrt {{{\rm{x}}^{\rm{3}}}{\rm{ + 1}}} {\rm{dx,}}\;\;\;{\rm{u = }}{{\rm{x}}^{\rm{3}}}{\rm{ + 1}}$$.

The solution of given integral$$\int {{x^2}} \sqrt {{x^3} + 1} dx$$ is $$\frac{2}{9} \cdot {({x^3} + 1)^{3/2}} + C$$.

See the step by step solution

## Step 1: Substitution method of evaluating integral

Substitution method is a method which transforms given integral into a simple form of integral by substituting the independent variable by others.

## Step 2: Applying substitution method

Let $$u = {x^3} + 1$$

\begin{aligned}{c}du &= 3{x^2}dx\\\frac{{du}}{3} &= {x^2}dx\end{aligned}

Substituting u and du in given function to get required results-

\begin{aligned}{c}\int {{x^2}} \cdot \sqrt {{x^3} + 1} dx &= \int {\sqrt u } \cdot \frac{{du}}{3}\\ &= \int {{u^{1/2}}} \cdot \frac{{du}}{3}\\ &= \frac{1}{3} \cdot \frac{{{u^{3/2}}}}{{3/2}} + C\\ &= \frac{2}{9} \cdot {u^{3/2}} + C\end{aligned}

Replace value of u with original value-

$$\int {{x^2}} \cdot \sqrt {{x^3} + 1} dx = \frac{2}{9} \cdot {({x^3} + 1)^{3/2}} + C$$

Therefore, the solution of given integral$$\int {{x^2}} \sqrt {{x^3} + 1} dx$$ is $$\frac{2}{9} \cdot {({x^3} + 1)^{3/2}} + C$$.