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

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

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

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

\frac{1}{2} arctan (2 x (x + 1)) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{( v - 1 ) v ( v - 2 )} d v

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

Apply u-substitution

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

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

Complete the square

w (w + 2) : (w + 1)^{2} - 1

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

Apply u-substitution

: \int \frac{1}{t t ^{2} - 1} d t

= \frac{1}{2} \cdot \int \frac{1}{t t ^{2} - 1} d t

Apply u-substitution

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

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

Use the common integral:

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

= \frac{1}{2} arctan (r)

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} arctan (2 x (x + 1)) + C

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