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

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

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

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

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

Decimal

1.06814 \dots

Solution steps

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

Expand

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

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

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

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

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

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\int_{0}^{5} (x - 1) d x

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\int_{0}^{4} (x + 1) d x

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