What is the integral of (200)/(x(100+x^4)) ?

The integral of (200)/(x(100+x^4)) is 50(1/25 ln(x)-1/100 ln|x^4+100|)+C

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integral of (200)/(x(100+x^4))

\int \frac{200}{x ( 100 + x ^{4} )} d x

50 (\frac{1}{25} ln (x) - \frac{1}{100} ln x^{4} + 100) + C

Solution steps

\int \frac{200}{x ( 100 + x ^{4} )} d x

Take the constant out:

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

= 200 \cdot \int \frac{1}{x ( 100 + x ^{4} )} d x

Apply u-substitution

= 200 \cdot \int \frac{1}{4 u ( 100 + u )} d u

Take the constant out:

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

= 200 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( 100 + u )} d u

Take the partial fraction of

\frac{1}{u ( 100 + u )} : \frac{1}{100 u} - \frac{1}{100 ( u + 100 )}

= 200 \cdot \frac{1}{4} \cdot \int \frac{1}{100 u} - \frac{1}{100 ( u + 100 )} 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

= 200 \cdot \frac{1}{4} (\int \frac{1}{100 u} d u - \int \frac{1}{100 ( u + 100 )} d u)

\int \frac{1}{100 u} d u = \frac{1}{100} ln ∣ u ∣

\int \frac{1}{100 ( u + 100 )} d u = \frac{1}{100} ln ∣ u + 100 ∣

= 200 \cdot \frac{1}{4} (\frac{1}{100} ln ∣ u ∣ - \frac{1}{100} ln ∣ u + 100 ∣)

Substitute back

u = x^{4}

= 200 \cdot \frac{1}{4} (\frac{1}{100} ln x^{4} - \frac{1}{100} ln x^{4} + 100)

Simplify

200 \cdot \frac{1}{4} (\frac{1}{100} ln x^{4} - \frac{1}{100} ln x^{4} + 100) : 50 (\frac{1}{25} ln (x) - \frac{1}{100} ln x^{4} + 100)

= 50 (\frac{1}{25} ln (x) - \frac{1}{100} ln x^{4} + 100)

Add a constant to the solution

= 50 (\frac{1}{25} ln (x) - \frac{1}{100} ln x^{4} + 100) + C

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What is the integral of (200)/(x(100+x^4)) ?
The integral of (200)/(x(100+x^4)) is 50(1/25 ln(x)-1/100 ln|x^4+100|)+C