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

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

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

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

arctan (2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{4 u ^{2} + 1} d u

= \int \frac{2}{4 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}{4 u ^{2} + 1} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} arctan (2 x)

Simplify

2 \cdot \frac{1}{2} arctan (2 x) : arctan (2 x)

= arctan (2 x)

Add a constant to the solution

= arctan (2 x) + C

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