What is the integral of x/(sqrt(36x^2+9)) ?

The integral of x/(sqrt(36x^2+9)) is 1/12 sqrt(4x^2+1)+C

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integral of x/(sqrt(36x^2+9))

\int \frac{x}{36 x ^{2} + 9} d x

\frac{1}{12} 4 x^{2} + 1 + C

Solution steps

\int \frac{x}{36 x ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{72 u} d u

= \int \frac{1}{72 u} d u

Take the constant out:

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

= \frac{1}{72} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{72} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 36 x^{2} + 9

= \frac{1}{72} \cdot 2 (36 x^{2} + 9)^{\frac{1}{2}}

Simplify

\frac{1}{72} \cdot 2 (36 x^{2} + 9)^{\frac{1}{2}} : \frac{1}{12} 4 x^{2} + 1

= \frac{1}{12} 4 x^{2} + 1

Add a constant to the solution

= \frac{1}{12} 4 x^{2} + 1 + C

x \to 0 + lim ((1 + 3 x)^{\frac{1}{x}})

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

\int \frac{ln ( x + 1 )}{( x + 1 )} d x

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

\frac{1}{s ^{2} + 6 s + 34}

What is the integral of x/(sqrt(36x^2+9)) ?
The integral of x/(sqrt(36x^2+9)) is 1/12 sqrt(4x^2+1)+C