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

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

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

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

ln (2) + \frac{3}{8}

Decimal

1.06814 \dots

Solution steps

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

Expand

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

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

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

\int_{1}^{2} \frac{1}{m} d m = ln (2)

\int_{1}^{2} \frac{1}{m ^{3}} d m = \frac{3}{8}

= ln (2) + \frac{3}{8}

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