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

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

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

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

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

Decimal

1.06814 \dots

Solution steps

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

Expand

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

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

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

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

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

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

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