What is the integral of 1/(2x^2+5x-7) ?

The integral of 1/(2x^2+5x-7) is 1/9 ln|x-1|-1/9 ln|2x+7|+C

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integral of 1/(2x^2+5x-7)

\int \frac{1}{2 x ^{2} + 5 x - 7} d x

\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ 2 x + 7 ∣ + C

Solution steps

\int \frac{1}{2 x ^{2} + 5 x - 7} d x

Take the partial fraction of

\frac{1}{2 x ^{2} + 5 x - 7} : \frac{1}{9 ( x - 1 )} - \frac{2}{9 ( 2 x + 7 )}

= \int \frac{1}{9 ( x - 1 )} - \frac{2}{9 ( 2 x + 7 )} 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}{9 ( x - 1 )} d x - \int \frac{2}{9 ( 2 x + 7 )} d x

\int \frac{1}{9 ( x - 1 )} d x = \frac{1}{9} ln ∣ x - 1 ∣

\int \frac{2}{9 ( 2 x + 7 )} d x = \frac{1}{9} ln ∣ 2 x + 7 ∣

= \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ 2 x + 7 ∣

Add a constant to the solution

= \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ 2 x + 7 ∣ + C

\frac{\partial}{\partial y} (z e^{x yz})

x \to 4 + lim (\frac{1}{( x + 4 ) ^{3}})

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

\frac{\partial}{\partial x} (15 - 5 (x + 3)^{2})

\int (y - y) d y

What is the integral of 1/(2x^2+5x-7) ?
The integral of 1/(2x^2+5x-7) is 1/9 ln|x-1|-1/9 ln|2x+7|+C