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

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

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integral of ((x-1))/((x^2+3x-4))

integral

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

ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{x - 1}{x ^{2} + 3 x - 4} d x

Expand

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

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{x}{x ^{2} + 3 x - 4} d x - \int \frac{1}{x ^{2} + 3 x - 4} d x

\int \frac{x}{x ^{2} + 3 x - 4} d x = \frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣

\int \frac{1}{x ^{2} + 3 x - 4} d x = \frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣

= \frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣ - (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣)

Simplify

\frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣ - (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣) : ln ∣ x + 4 ∣

= ln ∣ x + 4 ∣

Add a constant to the solution

= ln ∣ x + 4 ∣ + C

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

\frac{z}{( 1 - z ^{2} ) ^{\frac{1}{2}}}

3 x \cdot e^{1.5 x^{2}}

e^{x} \cdot cos^{2} (e^{x})

\frac{e ^{- 1}}{x ^{3}}

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