What is the integral of (3x)/(x^2+2x-8) ?

The integral of (3x)/(x^2+2x-8) is 3(1/2 ln|x^2+2x-8|+1/6 (ln| 1/3 (x+1)+1|-ln| 1/3 (x+1)-1|))+C

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integral of (3x)/(x^2+2x-8)

\int \frac{3 x}{x ^{2} + 2 x - 8} d x

3 (\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1)) + C

Solution steps

\int \frac{3 x}{x ^{2} + 2 x - 8} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} + 2 x - 8} d x

Complete the square

x^{2} + 2 x - 8 : (x + 1)^{2} - 9

= 3 \cdot \int \frac{x}{( x + 1 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{2} - 9} d u

= 3 \cdot \int \frac{u - 1}{u ^{2} - 9} d u

Expand

\frac{u - 1}{u ^{2} - 9} : \frac{u}{u ^{2} - 9} - \frac{1}{u ^{2} - 9}

Apply the Sum Rule:

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

= 3 (\int \frac{u}{u ^{2} - 9} d u - \int \frac{1}{u ^{2} - 9} d u)

\int \frac{u}{u ^{2} - 9} d u = \frac{1}{2} ln u^{2} - 9

\int \frac{1}{u ^{2} - 9} d u = - \frac{1}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1)

= 3 (\frac{1}{2} ln u^{2} - 9 - (- \frac{1}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1)))

Substitute back

u = x + 1

= 3 (\frac{1}{2} ln (x + 1)^{2} - 9 - (- \frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1)))

Simplify

3 (\frac{1}{2} ln (x + 1)^{2} - 9 - (- \frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1))) : 3 (\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1))

= 3 (\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1))

Add a constant to the solution

= 3 (\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1)) + C

\frac{d}{d x} (6 x + 3)

x \to 3 lim (\frac{x ^{2} + 1}{x + 3})

y^{^{'}} = y, y (1) = 1

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

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

What is the integral of (3x)/(x^2+2x-8) ?
The integral of (3x)/(x^2+2x-8) is 3(1/2 ln|x^2+2x-8|+1/6 (ln| 1/3 (x+1)+1|-ln| 1/3 (x+1)-1|))+C