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

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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

= \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

: 2 u^{\frac{1}{2}}

= \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

\int_{- 1}^{5} (∣ x - 2 ∣) d x

f (x) = 9^{x} + e^{x}

x^{3}, - 2

tan (x)

n = 1 \sum \infty \frac{n !}{e ^{n}}

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