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

The integral from 1 to 2 of 1/(x^5+x^3) is 3/8+(ln(5)-3ln(2))/2

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integral from 1 to 2 of 1/(x^5+x^3)

\int_{1}^{2} \frac{1}{x ^{5} + x ^{3}} d x

\frac{3}{8} + \frac{ln ( 5 ) - 3 ln ( 2 )}{2}

Decimal

0.13999 \dots

Solution steps

\int_{1}^{2} \frac{1}{x ^{5} + x ^{3}} d x

Take the partial fraction of

\frac{1}{x ^{5} + x ^{3}} : - \frac{1}{x} + \frac{1}{x ^{3}} + \frac{x}{x ^{2} + 1}

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

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

= - ln (2) + \frac{3}{8} + \frac{1}{2} (ln (5) - ln (2))

Simplify

- ln (2) + \frac{3}{8} + \frac{1}{2} (ln (5) - ln (2)) : \frac{3}{8} + \frac{ln ( 5 ) - 3 ln ( 2 )}{2}

= \frac{3}{8} + \frac{ln ( 5 ) - 3 ln ( 2 )}{2}

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What is the integral from 1 to 2 of 1/(x^5+x^3) ?
The integral from 1 to 2 of 1/(x^5+x^3) is 3/8+(ln(5)-3ln(2))/2