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

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

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integral of (t^2+1)/t

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

\frac{t ^{2}}{2} + ln ∣ t ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{t} d t = ln ∣ t ∣

= \frac{t ^{2}}{2} + ln ∣ t ∣

Add a constant to the solution

= \frac{t ^{2}}{2} + ln ∣ t ∣ + C

\frac{d}{d x} (\frac{( x ^{2} - 1 )}{x ^{2} + 1})

\frac{d}{d x} (\frac{cos ( x )}{sec ( x )})

\int \frac{e ^{2 x}}{1 - e ^{x}} d x

\frac{d}{d x} (ln (\frac{x + 2}{x - 1}))

y = \frac{x + 2 ^{x}}{cos ( π x )}

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