What is the integral of x/(x^2+4x-5) ?

The integral of x/(x^2+4x-5) is 1/6 ln|x-1|+5/6 ln|x+5|+C

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integral of x/(x^2+4x-5)

\int \frac{x}{x ^{2} + 4 x - 5} d x

\frac{1}{6} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{x}{x ^{2} + 4 x - 5} d x

Take the partial fraction of

\frac{x}{x ^{2} + 4 x - 5} : \frac{1}{6 ( x - 1 )} + \frac{5}{6 ( x + 5 )}

= \int \frac{1}{6 ( x - 1 )} + \frac{5}{6 ( x + 5 )} 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}{6 ( x - 1 )} d x + \int \frac{5}{6 ( x + 5 )} d x

\int \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln ∣ x - 1 ∣

\int \frac{5}{6 ( x + 5 )} d x = \frac{5}{6} ln ∣ x + 5 ∣

= \frac{1}{6} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 5 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 5 ∣ + C

\frac{12}{s}

\int_{0}^{3} x - 1 d x

f (x) = \frac{1}{x}

\int_{- 2}^{3} \frac{11}{x ^{4}} d x

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

What is the integral of x/(x^2+4x-5) ?
The integral of x/(x^2+4x-5) is 1/6 ln|x-1|+5/6 ln|x+5|+C