What is the integral of 1/((t+4)(t-1)) ?

The integral of 1/((t+4)(t-1)) is -1/5 ln|t+4|+1/5 ln|t-1|+C

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integral of 1/((t+4)(t-1))

\int \frac{1}{( t + 4 ) ( t - 1 )} d t

- \frac{1}{5} ln ∣ t + 4 ∣ + \frac{1}{5} ln ∣ t - 1 ∣ + C

Solution steps

\int \frac{1}{( t + 4 ) ( t - 1 )} d t

Take the partial fraction of

\frac{1}{( t + 4 ) ( t - 1 )} : - \frac{1}{5 ( t + 4 )} + \frac{1}{5 ( t - 1 )}

= \int - \frac{1}{5 ( t + 4 )} + \frac{1}{5 ( t - 1 )} d t

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}{5 ( t + 4 )} d t + \int \frac{1}{5 ( t - 1 )} d t

\int \frac{1}{5 ( t + 4 )} d t = \frac{1}{5} ln ∣ t + 4 ∣

\int \frac{1}{5 ( t - 1 )} d t = \frac{1}{5} ln ∣ t - 1 ∣

= - \frac{1}{5} ln ∣ t + 4 ∣ + \frac{1}{5} ln ∣ t - 1 ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ t + 4 ∣ + \frac{1}{5} ln ∣ t - 1 ∣ + C

(x y - y^{2}) d x - x^{2} d y = 0

\frac{\partial}{\partial x} (- 5 y^{2} sin (x y^{2}))

\frac{\partial}{\partial y} (20 (y - x^{2}))

x \to 1 lim (\frac{x ^{2} + 1}{x + 5})

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

What is the integral of 1/((t+4)(t-1)) ?
The integral of 1/((t+4)(t-1)) is -1/5 ln|t+4|+1/5 ln|t-1|+C