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Q3E

Expert-verifiedFound in: Page 306

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 \(\frac{2}{9} \cdot {({x^3} + 1)^{3/2}} + C\).

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

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 \(\frac{2}{9} \cdot {({x^3} + 1)^{3/2}} + C\).

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