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

The integral from 1 to 2 of 1/(x(x^2+1)) is (-ln(5)+3ln(2))/2

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

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

\frac{- ln ( 5 ) + 3 ln ( 2 )}{2}

Decimal

0.23500 \dots

Solution steps

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

Take the partial fraction of

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

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

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

\int_{1}^{2} \frac{x}{x ^{2} + 1} d x = \frac{1}{2} (ln (5) - ln (2))

= ln (2) - \frac{1}{2} (ln (5) - ln (2))

Simplify

ln (2) - \frac{1}{2} (ln (5) - ln (2)) : \frac{- ln ( 5 ) + 3 ln ( 2 )}{2}

= \frac{- ln ( 5 ) + 3 ln ( 2 )}{2}

What is the integral from 1 to 2 of 1/(x(x^2+1)) ?
The integral from 1 to 2 of 1/(x(x^2+1)) is (-ln(5)+3ln(2))/2