What is the integral of 3/(x^2+3x-10) ?

The integral of 3/(x^2+3x-10) is 3(1/7 ln|x-2|-1/7 ln|x+5|)+C

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integral of 3/(x^2+3x-10)

\int \frac{3}{x ^{2} + 3 x - 10} d x

3 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x + 5 ∣) + C

Solution steps

\int \frac{3}{x ^{2} + 3 x - 10} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{x ^{2} + 3 x - 10} d x

Take the partial fraction of

\frac{1}{x ^{2} + 3 x - 10} : \frac{1}{7 ( x - 2 )} - \frac{1}{7 ( x + 5 )}

= 3 \cdot \int \frac{1}{7 ( x - 2 )} - \frac{1}{7 ( x + 5 )} 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

= 3 (\int \frac{1}{7 ( x - 2 )} d x - \int \frac{1}{7 ( x + 5 )} d x)

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

\int \frac{1}{7 ( x + 5 )} d x = \frac{1}{7} ln ∣ x + 5 ∣

= 3 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x + 5 ∣)

Add a constant to the solution

= 3 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x + 5 ∣) + C

\frac{d y}{d x} + y = e^{2 x}

y = x e^{x^{2}}

\frac{d}{d x} (\frac{3 x - 9}{x + 5})

\int (π - x) \cdot cos (n x) d x

\int \frac{x ^{2}}{( x ^{3} - 5 ) ^{2}} d x

What is the integral of 3/(x^2+3x-10) ?
The integral of 3/(x^2+3x-10) is 3(1/7 ln|x-2|-1/7 ln|x+5|)+C