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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{t} d t = ln ∣ t ∣

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

= ln ∣ t ∣ - \frac{1}{2} ln t^{2} + 1

Add a constant to the solution

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

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