What is the integral of x/(sqrt(6x^2+5)) ?

The integral of x/(sqrt(6x^2+5)) is 1/6 sqrt(6x^2+5)+C

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integral of x/(sqrt(6x^2+5))

\int \frac{x}{6 x ^{2} + 5} d x

\frac{1}{6} 6 x^{2} + 5 + C

Solution steps

\int \frac{x}{6 x ^{2} + 5} d x

Apply u-substitution

: \int \frac{1}{12 u} d u

= \int \frac{1}{12 u} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{12} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 6 x^{2} + 5

= \frac{1}{12} \cdot 2 (6 x^{2} + 5)^{\frac{1}{2}}

Simplify

\frac{1}{12} \cdot 2 (6 x^{2} + 5)^{\frac{1}{2}} : \frac{1}{6} 6 x^{2} + 5

= \frac{1}{6} 6 x^{2} + 5

Add a constant to the solution

= \frac{1}{6} 6 x^{2} + 5 + C

f (x) = x^{4} e^{2 x}

4 x + \frac{16}{x ^{2}}

\int_{0}^{1} x^{2} e^{x^{3}} d x

\frac{d y}{d x} = - 0.5 y + x

\int \frac{x}{3} d x

What is the integral of x/(sqrt(6x^2+5)) ?
The integral of x/(sqrt(6x^2+5)) is 1/6 sqrt(6x^2+5)+C