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

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

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

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

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

Solution steps

\int \frac{4 x}{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

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

f (x) = tan (x), at x = 0

y^{^{'''}} + y^{^{''}} - 8 y^{^{'}} - 12 y = 0

(t (x) - t (x)^{3}) v^{'} + 2 v = 0

\frac{\partial}{\partial x} (x y e^{x - y})

\frac{7}{x ^{6}}

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