What is the integral of 1/(xsqrt(9x-1)) ?

The integral of 1/(xsqrt(9x-1)) is 2arctan(sqrt(9x-1))+C

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integral of 1/(xsqrt(9x-1))

\int \frac{1}{x 9 x - 1} d x

2 arctan (9 x - 1) + C

Solution steps

\int \frac{1}{x 9 x - 1} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = 9 x - 1

= 2 arctan (9 x - 1)

Add a constant to the solution

= 2 arctan (9 x - 1) + C

\int (\frac{cos ( x )}{sin ( 2 x )}) d x

\frac{x + 1770}{30 ( x + 2 )}

\frac{d}{d x} (2 x (3 x - 2)^{4})

\int \frac{t ^{5}}{t ^{6} + 1} d t

\frac{d}{d x} (tan (\frac{x}{y}))

What is the integral of 1/(xsqrt(9x-1)) ?
The integral of 1/(xsqrt(9x-1)) is 2arctan(sqrt(9x-1))+C