What is the integral of x^3\sqrt[3]{x^2+1} ?

The integral of x^3\sqrt[3]{x^2+1} is 1/2 (3/7 (x^2+1)^{7/3}-3/4 (x^2+1)^{4/3})+C

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integral of x^3\sqrt[3]{x^2+1}

\frac{1}{2} (\frac{3}{7} (x^{2} + 1)^{\frac{7}{3}} - \frac{3}{4} (x^{2} + 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}{2} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = x^{2} + 1

= \frac{1}{2} (\frac{3}{7} (x^{2} + 1)^{\frac{7}{3}} - \frac{3}{4} (x^{2} + 1)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{2} (\frac{3}{7} (x^{2} + 1)^{\frac{7}{3}} - \frac{3}{4} (x^{2} + 1)^{\frac{4}{3}}) + C

y^{^{'}} = - y t + 0.1 y^{^{'}} t + 4

\int \frac{( 3 x ^{2} + 4 )}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\int 8 x (x^{7} - \frac{1}{8}) d x

x \to \frac{π}{3} lim (sin (x))

\frac{d}{d x} (x^{2} a + a^{2} b + b^{2} x)

What is the integral of x^3\sqrt[3]{x^2+1} ?
The integral of x^3\sqrt[3]{x^2+1} is 1/2 (3/7 (x^2+1)^{7/3}-3/4 (x^2+1)^{4/3})+C