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

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

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

\int \frac{1}{( t + 2 ) ^{2} ( t + 1 )} d t

- ln ∣ t + 2 ∣ + \frac{1}{t + 2} + ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{1}{( t + 2 ) ^{2} ( t + 1 )} d t

Take the partial fraction of

\frac{1}{( t + 2 ) ^{2} ( t + 1 )} : - \frac{1}{t + 2} - \frac{1}{( t + 2 ) ^{2}} + \frac{1}{t + 1}

= \int - \frac{1}{t + 2} - \frac{1}{( t + 2 ) ^{2}} + \frac{1}{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}{t + 2} d t - \int \frac{1}{( t + 2 ) ^{2}} d t + \int \frac{1}{t + 1} d t

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

\int \frac{1}{( t + 2 ) ^{2}} d t = - \frac{1}{t + 2}

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

= - ln ∣ t + 2 ∣ - (- \frac{1}{t + 2}) + ln ∣ t + 1 ∣

Simplify

= - ln ∣ t + 2 ∣ + \frac{1}{t + 2} + ln ∣ t + 1 ∣

Add a constant to the solution

= - ln ∣ t + 2 ∣ + \frac{1}{t + 2} + ln ∣ t + 1 ∣ + C

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