What is the integral of (x^3)/(sqrt(x^4+9)) ?

The integral of (x^3)/(sqrt(x^4+9)) is 1/2 sqrt(x^4+9)+C

integral of (x^3)/(sqrt(x^4+9))

\int \frac{x ^{3}}{x ^{4} + 9} d x

\frac{1}{2} x^{4} + 9 + C

Solution steps

\int \frac{x ^{3}}{x ^{4} + 9} d x

Apply u-substitution

= \int \frac{1}{4 u} d u

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = x^{4} + 9

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

Simplify

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

= \frac{1}{2} x^{4} + 9

Add a constant to the solution

= \frac{1}{2} x^{4} + 9 + C

d er i v a t i v e f (x) = 9^{x} + e^{x}

t a y l or x^{3}, - 2

d er i v a t i v e tan (x)

t an g e n t f (x) = \frac{2}{x}, (- 8, - \frac{1}{4})

\frac{\partial}{\partial y} (\frac{x - y}{x + 2 y})

What is the integral of (x^3)/(sqrt(x^4+9)) ?
The integral of (x^3)/(sqrt(x^4+9)) is 1/2 sqrt(x^4+9)+C