What is the integral of 1/(x^2+6x-7) ?

The integral of 1/(x^2+6x-7) is 1/8 ln|x-1|-1/8 ln|x+7|+C

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integral of 1/(x^2+6x-7)

\int \frac{1}{x ^{2} + 6 x - 7} d x

\frac{1}{8} ln ∣ x - 1 ∣ - \frac{1}{8} ln ∣ x + 7 ∣ + C

Solution steps

\int \frac{1}{x ^{2} + 6 x - 7} d x

Take the partial fraction of

\frac{1}{x ^{2} + 6 x - 7} : \frac{1}{8 ( x - 1 )} - \frac{1}{8 ( x + 7 )}

= \int \frac{1}{8 ( x - 1 )} - \frac{1}{8 ( x + 7 )} d x

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{1}{8 ( x - 1 )} d x - \int \frac{1}{8 ( x + 7 )} d x

\int \frac{1}{8 ( x - 1 )} d x = \frac{1}{8} ln ∣ x - 1 ∣

\int \frac{1}{8 ( x + 7 )} d x = \frac{1}{8} ln ∣ x + 7 ∣

= \frac{1}{8} ln ∣ x - 1 ∣ - \frac{1}{8} ln ∣ x + 7 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ x - 1 ∣ - \frac{1}{8} ln ∣ x + 7 ∣ + C

\int \frac{1}{sec ( u )} d u

3 y^{^{'}} + \frac{2 y}{x} = \frac{( y ^{4} )}{x ^{4}}

2 x^{3} + 3 x^{2} - 12 x + 7

y = x^{2}, y = 9, [- 3, 3]

x^{3} cos (x^{4})

What is the integral of 1/(x^2+6x-7) ?
The integral of 1/(x^2+6x-7) is 1/8 ln|x-1|-1/8 ln|x+7|+C