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

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

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

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

arctan (36 x^{2} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

= \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= arctan (u)

Substitute back

u = 36 x^{2} - 1

= arctan (36 x^{2} - 1)

Add a constant to the solution

= arctan (36 x^{2} - 1) + C

\int 6 x^{2} e^{6 x} d x

y = x^{3} + x^{2} - 6 x, y = 6 x

e^{(3 t)} (4 cos (3 t) - \frac{10}{3} sin (3 t))

x \to - 1 + lim (\frac{x}{∣ x + 1 ∣})

y, y = x e^{y}

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