What is the integral of x/(sqrt(36x^2-1)) ?

The integral of x/(sqrt(36x^2-1)) is 1/36 sqrt(36x^2-1)+C

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integral of x/(sqrt(36x^2-1))

\int \frac{x}{36 x ^{2} - 1} d x

\frac{1}{36} 36 x^{2} - 1 + C

Solution steps

\int \frac{x}{36 x ^{2} - 1} 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} - 1

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

Simplify

\frac{1}{72} \cdot 2 (36 x^{2} - 1)^{\frac{1}{2}} : \frac{1}{36} 36 x^{2} - 1

= \frac{1}{36} 36 x^{2} - 1

Add a constant to the solution

= \frac{1}{36} 36 x^{2} - 1 + C

f (x) = 6 e^{x} + \frac{2}{3 x}

\frac{\partial}{\partial x} (ln (x^{7} + y^{2}))

\int \frac{4 x}{1 + x ^{2}} d x

x \to 1 lim (\frac{x}{x ^{2} - 2 x})

f (x) = x^{3} - 3 x + 2

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