What is the integral from 1 to 3 of (2t^2+1)/t ?

The integral from 1 to 3 of (2t^2+1)/t is 8+ln(3)

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integral from 1 to 3 of (2t^2+1)/t

\int_{1}^{3} \frac{2 t ^{2} + 1}{t} d t

8 + ln (3)

Decimal Notation

9.09861 \dots

Solution steps

\int_{1}^{3} \frac{2 t ^{2} + 1}{t} d t

Expand

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

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

\int_{1}^{3} 2 t d t = 8

\int_{1}^{3} \frac{1}{t} d t = ln (3)

= 8 + ln (3)

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What is the integral from 1 to 3 of (2t^2+1)/t ?
The integral from 1 to 3 of (2t^2+1)/t is 8+ln(3)