What is the integral of \sqrt[3]{x^3+1}x^5 ?

The integral of \sqrt[3]{x^3+1}x^5 is 1/28 (4(x^3+1)^{7/3}-7(x^3+1)^{4/3})+C

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integral of \sqrt[3]{x^3+1}x^5

\frac{1}{28} (4 (x^{3} + 1)^{\frac{7}{3}} - 7 (x^{3} + 1)^{\frac{4}{3}}) + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

= \frac{1}{3} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = x^{3} + 1

= \frac{1}{3} (\frac{3}{7} (x^{3} + 1)^{\frac{7}{3}} - \frac{3}{4} (x^{3} + 1)^{\frac{4}{3}})

Simplify

\frac{1}{3} (\frac{3}{7} (x^{3} + 1)^{\frac{7}{3}} - \frac{3}{4} (x^{3} + 1)^{\frac{4}{3}}) : \frac{1}{28} (4 (x^{3} + 1)^{\frac{7}{3}} - 7 (x^{3} + 1)^{\frac{4}{3}})

= \frac{1}{28} (4 (x^{3} + 1)^{\frac{7}{3}} - 7 (x^{3} + 1)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{28} (4 (x^{3} + 1)^{\frac{7}{3}} - 7 (x^{3} + 1)^{\frac{4}{3}}) + C

\int \frac{1 - sin ( x )}{cos ( x )} d x

\frac{d}{d x} (x + e^{- 2 x})

x^{2} y d x + (x^{3} + 3 x y^{2}) d y = 0

t y^{^{'}} + (1 + t) y = t

y^{^{''}} + 7 y^{^{'}} + 12 y = 2 e^{t}

What is the integral of \sqrt[3]{x^3+1}x^5 ?
The integral of \sqrt[3]{x^3+1}x^5 is 1/28 (4(x^3+1)^{7/3}-7(x^3+1)^{4/3})+C