What is the integral of (sqrt(x))/(x^3-1) ?

The integral of (sqrt(x))/(x^3-1) is -1/3 (ln|x^{3/2}+1|-ln|x^{3/2}-1|)+C

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integral of (sqrt(x))/(x^3-1)

\int \frac{x}{x ^{3} - 1} d x

- \frac{1}{3} (ln x^{\frac{3}{2}} + 1 - ln x^{\frac{3}{2}} - 1) + C

Solution steps

\int \frac{x}{x ^{3} - 1} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{6} - 1} d u

= \int \frac{2 u ^{2}}{u ^{6} - 1} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{6} - 1} d u

Apply u-substitution

: \int \frac{1}{3 ( v ^{2} - 1 )} d v

= 2 \cdot \int \frac{1}{3 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 \cdot \frac{1}{3} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 2 \cdot \frac{1}{3} - \frac{ln ( x ) ^{3} + 1}{2} - \frac{ln ( x ) ^{3} - 1}{2}

Simplify

2 \cdot \frac{1}{3} - \frac{ln ( x ) ^{3} + 1}{2} - \frac{ln ( x ) ^{3} - 1}{2} : - \frac{1}{3} (ln x^{\frac{3}{2}} + 1 - ln x^{\frac{3}{2}} - 1)

= - \frac{1}{3} (ln x^{\frac{3}{2}} + 1 - ln x^{\frac{3}{2}} - 1)

Add a constant to the solution

= - \frac{1}{3} (ln x^{\frac{3}{2}} + 1 - ln x^{\frac{3}{2}} - 1) + C

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What is the integral of (sqrt(x))/(x^3-1) ?
The integral of (sqrt(x))/(x^3-1) is -1/3 (ln|x^{3/2}+1|-ln|x^{3/2}-1|)+C