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

The integral of 1/(4x^2+4x-3) is -1/8 (ln|x+3/2 |-ln|x-1/2 |)+C

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

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

- \frac{1}{8} (ln x + \frac{3}{2} - ln x - \frac{1}{2}) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{4} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = x + \frac{1}{2}

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

Simplify

\frac{1}{4} (- (\frac{ln x + \frac{1}{2} + 1}{2} - \frac{ln x + \frac{1}{2} - 1}{2})) : - \frac{1}{8} (ln x + \frac{3}{2} - ln x - \frac{1}{2})

= - \frac{1}{8} (ln x + \frac{3}{2} - ln x - \frac{1}{2})

Add a constant to the solution

= - \frac{1}{8} (ln x + \frac{3}{2} - ln x - \frac{1}{2}) + C

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What is the integral of 1/(4x^2+4x-3) ?
The integral of 1/(4x^2+4x-3) is -1/8 (ln|x+3/2 |-ln|x-1/2 |)+C