What is the integral of 1/(x^{-1/2)-x} ?

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

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integral of 1/(x^{-1/2)-x}

\int \frac{1}{x ^{- \frac{1}{2}} - x} d x

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

Solution steps

\int \frac{1}{x ^{- \frac{1}{2}} - x} d x

Factor

\frac{1}{x ^{- \frac{1}{2}} - x}

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

Simplify

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

Apply u-substitution

: \int - \frac{2}{3 u} d u

= \int - \frac{2}{3 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{2}{3} ln ∣ u ∣

Substitute back

u = 1 - x^{\frac{3}{2}}

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

Add a constant to the solution

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

\frac{d y}{d x} + 2 y = e^{x} y^{- 4}

\frac{d y}{d x} = 2 x y + x y^{2}

n = 0 \sum \infty 2 + (0.86)^{n}

\int (e^{2 x} - e^{- 2 x})^{2} d x

\frac{d y}{d t} = k y

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