Evaluate the integral by making the given substitution. ∫x2x3+1dx, u=x3+1

∫x2x3+1dx Let u=x3+1⇒du=3x2dx⇒13du=x3dx Apply the substitution ∫x2x3+1dx=∫x3+1⏞ux2dx⏞13du=∫u(13)du 13∫u12du Integrate, apply ∫undu=un+1n+1+C, so 13∫u12du=13(u3232)+C 2u329+C Back - substitute u=x3+1 2(x3+1)329+C Results: 2(x3+1)329+C

Resolved: "Evaluate the integral by making the given substitution...."

vestirme4

Answered question

2021-10-17

Evaluate the integral by making the given substitution.

\int x^{2} \sqrt{x^{3} + 1} d x, u = x^{3} + 1

Answer & Explanation

Nathanael Webber

Skilled2021-10-18Added 117 answers

\int x^{2} \sqrt{x^{3} + 1} d x

Let

u = x^{3} + 1 \Rightarrow d u = 3 x^{2} d x \Rightarrow \frac{1}{3} d u = x^{3} d x

Apply the substitution

\int x^{2} \sqrt{x^{3} + 1} d x = \int \overset{\sqrt{u}}{\overset{⏞}{\sqrt{x^{3} + 1}}} \overset{\frac{1}{3} d u}{\overset{⏞}{x^{2} d x}} = \int \sqrt{u} (\frac{1}{3}) d u

\frac{1}{3} \int u^{\frac{1}{2}} d u

Integrate, apply

\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C

, so

\frac{1}{3} \int u^{\frac{1}{2}} d u = \frac{1}{3} (\frac{u^{\frac{3}{2}}}{\frac{3}{2}}) + C

\frac{2 u^{\frac{3}{2}}}{9} + C

Back - substitute

u = x^{3} + 1

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

Results:

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

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