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

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

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

integral

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

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

Solution steps

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

Expand

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

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

9 cos^{2} (x)

x lo g_{10} (x^{2} + 5)

f (x) = (x^{2} + x + 1)^{\frac{1}{2}}

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

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

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