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Popular >

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

Solution

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

Solution

\frac{1}{2} ln 4 x^{2} + 4 x - 4 + \frac{5}{2} ln \frac{1}{5} (2 x + 1) + 1 - \frac{5}{2} ln \frac{1}{5} (2 x + 1) - 1 + C

Solution steps

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

Expand

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

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

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

\int \frac{2}{x ^{2} + x - 1} d x = - \frac{2}{5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1)

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 4 + \frac{1}{4 5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1)) - (- \frac{2}{5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1))

Simplify

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

= \frac{1}{2} ln 4 x^{2} + 4 x - 4 + \frac{5}{2} ln \frac{1}{5} (2 x + 1) + 1 - \frac{5}{2} ln \frac{1}{5} (2 x + 1) - 1

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} + 4 x - 4 + \frac{5}{2} ln \frac{1}{5} (2 x + 1) + 1 - \frac{5}{2} ln \frac{1}{5} (2 x + 1) - 1 + C

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

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\int sin^{5} (4 x) d x

limit as x approaches 9 of 1/(x^2)

x \to 9 lim (\frac{1}{x ^{2}})

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

integral of cos^2(2ax)

\int cos^{2} (2 a x) d x

Frequently Asked Questions (FAQ)

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