What is the integral of ((x+3))/(x^2+8x-9) ?

The integral of ((x+3))/(x^2+8x-9) is ln|x+9|+4(1/10 ln|x-1|-1/10 ln|x+9|)+C

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integral of ((x+3))/(x^2+8x-9)

\int \frac{( x + 3 )}{x ^{2} + 8 x - 9} d x

ln ∣ x + 9 ∣ + 4 (\frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x + 9 ∣) + C

Solution steps

\int \frac{x + 3}{x ^{2} + 8 x - 9} d x

Factor

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

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

Apply u-substitution

: \int \frac{u + 4}{u ( u + 10 )} d u

= \int \frac{u + 4}{u ( u + 10 )} d u

Expand

\frac{u + 4}{u ( u + 10 )} : \frac{u}{u ( u + 10 )} + \frac{4}{u ( u + 10 )}

Apply the Sum Rule:

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

= \int \frac{u}{u ( u + 10 )} d u + \int \frac{4}{u ( u + 10 )} d u

\int \frac{u}{u ( u + 10 )} d u = ln ∣ u + 10 ∣

\int \frac{4}{u ( u + 10 )} d u = 4 (\frac{1}{10} ln ∣ u ∣ - \frac{1}{10} ln ∣ u + 10 ∣)

= ln ∣ u + 10 ∣ + 4 (\frac{1}{10} ln ∣ u ∣ - \frac{1}{10} ln ∣ u + 10 ∣)

Substitute back

u = x - 1

= ln ∣ x - 1 + 10 ∣ + 4 (\frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x - 1 + 10 ∣)

Simplify

ln ∣ x - 1 + 10 ∣ + 4 (\frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x - 1 + 10 ∣) : ln ∣ x + 9 ∣ + 4 (\frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x + 9 ∣)

= ln ∣ x + 9 ∣ + 4 (\frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x + 9 ∣)

Add a constant to the solution

= ln ∣ x + 9 ∣ + 4 (\frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x + 9 ∣) + C

x \to 2 - lim (\frac{x ^{2}}{x - 2})

n = 1 \sum \infty n^{3} (1 + n^{4})

f (x) = x \cdot e^{2 x - 4}, x = 2

\int \frac{( sin ( x ) )}{x} d x

(3, - 1), (4, 7)

What is the integral of ((x+3))/(x^2+8x-9) ?
The integral of ((x+3))/(x^2+8x-9) is ln|x+9|+4(1/10 ln|x-1|-1/10 ln|x+9|)+C