What is the integral of 7/((x^2+3x-10)) ?

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

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integral of 7/((x^2+3x-10))

\int \frac{7}{( x ^{2} + 3 x - 10 )} d x

ln ∣ x - 2 ∣ - ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{7}{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

= 7 \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 )}

= 7 \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

= 7 (\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 ∣

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

Simplify

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

= ln ∣ x - 2 ∣ - ln ∣ x + 5 ∣

Add a constant to the solution

= ln ∣ x - 2 ∣ - ln ∣ x + 5 ∣ + C

x \to 0 lim (x^{2} \cdot sin (\frac{1}{x ^{2}}))

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

tan (θ)

\int_{1}^{2} \int_{0}^{0} x y e^{x y^{2}} d x d y

\frac{\partial}{\partial x} (z)

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