What is the integral of 1/(xsqrt(4x-1)) ?

The integral of 1/(xsqrt(4x-1)) is 2arctan(sqrt(4x-1))+C

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integral of 1/(xsqrt(4x-1))

\int \frac{1}{x 4 x - 1} d x

2 arctan (4 x - 1) + C

Solution steps

\int \frac{1}{x 4 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 = 4 x - 1

= 2 arctan (4 x - 1)

Add a constant to the solution

= 2 arctan (4 x - 1) + C

\int (x + 1) e^{6 x^{2} + 12 x} d x

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

\frac{\partial}{\partial t} (t - 2 s)

What is the integral of 1/(xsqrt(4x-1)) ?
The integral of 1/(xsqrt(4x-1)) is 2arctan(sqrt(4x-1))+C