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

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

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

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

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

Solution steps

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

Simplify

(1 + \frac{1}{t}) \frac{1}{t ^{2}} : \frac{t + 1}{t ^{3}}

= \int \frac{t + 1}{t ^{3}} d t

Expand

\frac{t + 1}{t ^{3}} : \frac{1}{t ^{2}} + \frac{1}{t ^{3}}

= \int \frac{1}{t ^{2}} + \frac{1}{t ^{3}} 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 ^{3}} d t

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

\int \frac{1}{t ^{3}} d t = - \frac{1}{2 t ^{2}}

= - \frac{1}{t} - \frac{1}{2 t ^{2}}

Add a constant to the solution

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

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