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

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

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

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

x - ln ∣ x ∣ + C

Solution steps

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

Simplify

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

= \int \frac{x - 1}{x} d x

Expand

\frac{x - 1}{x} : 1 - \frac{1}{x}

= \int 1 - \frac{1}{x} 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 1 d x - \int \frac{1}{x} d x

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

= x - ln ∣ x ∣

Add a constant to the solution

= x - ln ∣ x ∣ + C

(x^{''})^{^{'}}

y^{^{''}} + a \cdot y^{^{'}} = b, y (0) = 0, y^{^{'}} (0) = 0

\frac{d y}{d x} = \frac{( 2 x + sec ^{2} ( x ) )}{2 y}, y (0) = - 6

\int \frac{3 x + 5}{x ^{2} + 1} d x

\int \frac{3 x ^{2} - 6 x - 4}{x ^{3} - 3 x ^{2} - 4 x + 12} d x

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