What is the integral of 1/((2x+1)sqrt(4x^2+4x-3)) ?

The integral of 1/((2x+1)sqrt(4x^2+4x-3)) is 1/4 arcsec(x+1/2)+C

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integral of 1/((2x+1)sqrt(4x^2+4x-3))

\int \frac{1}{( 2 x + 1 ) 4 x ^{2} + 4 x - 3} d x

\frac{1}{4} arcsec (x + \frac{1}{2}) + C

Solution steps

\int \frac{1}{( 2 x + 1 ) 4 x ^{2} + 4 x - 3} d x

Apply u-substitution

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

Apply Trigonometric Substitution

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (2 x + 1))

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (2 x + 1)) : \frac{1}{4} arcsec (x + \frac{1}{2})

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

Add a constant to the solution

= \frac{1}{4} arcsec (x + \frac{1}{2}) + C

What is the integral of 1/((2x+1)sqrt(4x^2+4x-3)) ?
The integral of 1/((2x+1)sqrt(4x^2+4x-3)) is 1/4 arcsec(x+1/2)+C