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

The integral of 1/(sqrt(x(4x-1))) is ln|2sqrt(x)+sqrt(4x-1)|+C

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

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

ln 2 x + 4 x - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ( u + 1 )} d u

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

Take the constant out:

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

u (u + 1) = u (u + 1)

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = ln v^{2} + 1 + v

= \frac{1}{2} \cdot 2 ln v^{2} + 1 + v

Substitute back

= \frac{1}{2} \cdot 2 ln (4 x - 1)^{2} + 1 + 4 x - 1

Simplify

\frac{1}{2} \cdot 2 ln (4 x - 1)^{2} + 1 + 4 x - 1 : ln 2 x + 4 x - 1

= ln 2 x + 4 x - 1

Add a constant to the solution

= ln 2 x + 4 x - 1 + C

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