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

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

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

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

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

Solution steps

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

Factor

x^{2} + 2 x - 3 : - (- x^{2} - 2 x + 3)

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

Apply Integral Substitution

= - \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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{6}}{6})

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

3 y^{^{''}} + 13 y^{^{'}} + 4 y = 0

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

\frac{d}{d x} (\frac{4}{3 x})

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