What is the integral of (5x)/(sqrt(3x^2+4)) ?

The integral of (5x)/(sqrt(3x^2+4)) is 5/3 sqrt(3x^2+4)+C

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integral of (5x)/(sqrt(3x^2+4))

\int \frac{5 x}{3 x ^{2} + 4} d x

\frac{5}{3} 3 x^{2} + 4 + C

Solution steps

\int \frac{5 x}{3 x ^{2} + 4} d x

Take the constant out:

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

= 5 \cdot \int \frac{x}{3 x ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= 5 \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 5 \cdot \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 3 x^{2} + 4

= 5 \cdot \frac{1}{6} \cdot 2 (3 x^{2} + 4)^{\frac{1}{2}}

Simplify

5 \cdot \frac{1}{6} \cdot 2 (3 x^{2} + 4)^{\frac{1}{2}} : \frac{5}{3} 3 x^{2} + 4

= \frac{5}{3} 3 x^{2} + 4

Add a constant to the solution

= \frac{5}{3} 3 x^{2} + 4 + C

n = 0 \sum \infty 5 (\frac{3}{2})^{n}

\int x^{2} 8 x + 9 d x

\frac{\partial}{\partial y} (\frac{y ^{3}}{x ^{2} + y ^{2}})

n = 0 \sum \infty \frac{3 n ^{9}}{n !}

\frac{d}{d x} (3 cos (x) + 3 sin (x))

What is the integral of (5x)/(sqrt(3x^2+4)) ?
The integral of (5x)/(sqrt(3x^2+4)) is 5/3 sqrt(3x^2+4)+C