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

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

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integral of 1/(2-t)+1/(2+t)

\int \frac{1}{2 - t} + \frac{1}{2 + t} d t

- ln ∣ 2 - t ∣ + ln ∣ 2 + t ∣ + C

Solution steps

\int \frac{1}{2 - t} + \frac{1}{2 + t} 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}{2 - t} d t + \int \frac{1}{2 + t} d t

\int \frac{1}{2 - t} d t = - ln ∣ 2 - t ∣

\int \frac{1}{2 + t} d t = ln ∣ 2 + t ∣

= - ln ∣ 2 - t ∣ + ln ∣ 2 + t ∣

Add a constant to the solution

= - ln ∣ 2 - t ∣ + ln ∣ 2 + t ∣ + C

\int_{- \infty}^{4} e^{- x} d x

y^{^{''}} + y^{^{'}} = sin^{2} (x)

x = - 2, x = 1, y = 11 x, y = x^{2} - 12

\frac{d}{d x} (400 - 5 x)

\int v (u + v^{2})^{4} d u

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