What is the integral of x\sqrt[3]{x+c} ?

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

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integral of x\sqrt[3]{x+c}

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

Solution steps

Apply u-substitution

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{3}{7} u^{\frac{7}{3}} - c \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = x + c

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

Add a constant to the solution

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

\int - 10 x^{4} y^{5} d x

ln (4 - x^{2})

\int \frac{1 - cos ^{2} ( x )}{sin ( x )} d x

y^{^{'}} = 2 x + y

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

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