What is the integral from 1 to 2 of 1/(x(1+x^4)) ?

The integral from 1 to 2 of 1/(x(1+x^4)) is (5ln(2)-ln(17))/4

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integral from 1 to 2 of 1/(x(1+x^4))

\int_{1}^{2} \frac{1}{x ( 1 + x ^{4} )} d x

\frac{5 ln ( 2 ) - ln ( 17 )}{4}

Decimal

0.15813 \dots

Solution steps

\int_{1}^{2} \frac{1}{x ( 1 + x ^{4} )} d x

Apply u-substitution

: \int_{2}^{17} \frac{1}{4 ( u - 1 ) u} d u

= \int_{2}^{17} \frac{1}{4 ( u - 1 ) u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{2}^{17} \frac{1}{( u - 1 ) u} d u

Take the partial fraction of

\frac{1}{( u - 1 ) u} : \frac{1}{u - 1} - \frac{1}{u}

= \frac{1}{4} \cdot \int_{2}^{17} \frac{1}{u - 1} - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int_{2}^{17} \frac{1}{u - 1} d u - \int_{2}^{17} \frac{1}{u} d u)

\int_{2}^{17} \frac{1}{u - 1} d u = 4 ln (2)

\int_{2}^{17} \frac{1}{u} d u = ln (17) - ln (2)

= \frac{1}{4} (4 ln (2) - (ln (17) - ln (2)))

Simplify

\frac{1}{4} (4 ln (2) - (ln (17) - ln (2))) : \frac{5 ln ( 2 ) - ln ( 17 )}{4}

= \frac{5 ln ( 2 ) - ln ( 17 )}{4}

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What is the integral from 1 to 2 of 1/(x(1+x^4)) ?
The integral from 1 to 2 of 1/(x(1+x^4)) is (5ln(2)-ln(17))/4