What is the integral of 1/((x+1/2)sqrt(4x^2+4x)) ?

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

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integral of 1/((x+1/2)sqrt(4x^2+4x))

\int \frac{1}{( x + \frac{1}{2} ) 4 x ^{2} + 4 x} d x

arctan (2 x (x + 1)) + C

Solution steps

\int \frac{1}{( x + \frac{1}{2} ) 4 x ^{2} + 4 x} d x

Complete the square

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

= \int \frac{1}{( x + \frac{1}{2} ) 4 ( x + \frac{1}{2} ) ^{2} - 1} d x

Apply u-substitution

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (v)

Substitute back

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

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

= arctan (2 x (x + 1))

Add a constant to the solution

= arctan (2 x (x + 1)) + C

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