What is the integral of 7x^2sqrt(x^3+15) ?

The integral of 7x^2sqrt(x^3+15) is 14/9 (x^3+15)^{3/2}+C

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integral of 7x^2sqrt(x^3+15)

\int 7 x^{2} x^{3} + 15 d x

\frac{14}{9} (x^{3} + 15)^{\frac{3}{2}} + C

Solution steps

\int 7 x^{2} x^{3} + 15 d x

Take the constant out:

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

= 7 \cdot \int x^{2} x^{3} + 15 d x

Apply u-substitution

: \int \frac{u}{3} d u

= 7 \cdot \int \frac{u}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 7 \cdot \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{3} + 15

= 7 \cdot \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 15)^{\frac{3}{2}}

Simplify

7 \cdot \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 15)^{\frac{3}{2}} : \frac{14}{9} (x^{3} + 15)^{\frac{3}{2}}

= \frac{14}{9} (x^{3} + 15)^{\frac{3}{2}}

Add a constant to the solution

= \frac{14}{9} (x^{3} + 15)^{\frac{3}{2}} + C

\frac{d y}{d x} = 3 x + y

9 x - 9 x

\frac{d}{d x} (ln (\frac{a ^{x}}{1 + a ^{x}}))

(1 + x) \frac{d y}{d x} - x y = x - x^{2}

\frac{\partial}{\partial x} (10 arccot (x))

What is the integral of 7x^2sqrt(x^3+15) ?
The integral of 7x^2sqrt(x^3+15) is 14/9 (x^3+15)^{3/2}+C