What is the integral of (3xsqrt(2x^2+9))/5 ?

The integral of (3xsqrt(2x^2+9))/5 is 1/10 (2x^2+9)^{3/2}+C

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integral of (3xsqrt(2x^2+9))/5

\int \frac{3 x 2 x ^{2} + 9}{5} d x

\frac{1}{10} (2 x^{2} + 9)^{\frac{3}{2}} + C

Solution steps

\int \frac{3 x 2 x ^{2} + 9}{5} d x

Take the constant out:

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

= \frac{3}{5} \cdot \int x 2 x^{2} + 9 d x

Apply u-substitution

: \int \frac{u}{4} d u

= \frac{3}{5} \cdot \int \frac{u}{4} d u

Take the constant out:

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

= \frac{3}{5} \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= \frac{3}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 x^{2} + 9

= \frac{3}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} (2 x^{2} + 9)^{\frac{3}{2}}

Simplify

\frac{3}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} (2 x^{2} + 9)^{\frac{3}{2}} : \frac{1}{10} (2 x^{2} + 9)^{\frac{3}{2}}

= \frac{1}{10} (2 x^{2} + 9)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{10} (2 x^{2} + 9)^{\frac{3}{2}} + C

\int_{0}^{7} y + 9 d y

f (x) = 2 sin (x), at x = \frac{π}{6}

y = x, y = \frac{1}{2} x, x = 9

\int cos^{3} (\frac{x}{14}) d x

\frac{\partial}{\partial x} (e^{5 x yz - 6})

What is the integral of (3xsqrt(2x^2+9))/5 ?
The integral of (3xsqrt(2x^2+9))/5 is 1/10 (2x^2+9)^{3/2}+C