What is the integral of 1/(2-x^{1/5)} ?

The integral of 1/(2-x^{1/5)} is 5(-1/4 x^{4/5}-2/3 x^{3/5}-2x^{2/5}-8x^{1/5}-16ln|-x^{1/5}+2|)+C

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integral of 1/(2-x^{1/5)}

\int \frac{1}{2 - x ^{\frac{1}{5}}} d x

5 (- \frac{1}{4} x^{\frac{4}{5}} - \frac{2}{3} x^{\frac{3}{5}} - 2 x^{\frac{2}{5}} - 8 x^{\frac{1}{5}} - 16 ln - x^{\frac{1}{5}} + 2) + C

Solution steps

\int \frac{1}{2 - x ^{\frac{1}{5}}} d x

Apply u-substitution

: \int \frac{5 u ^{4}}{2 - u} d u

= \int \frac{5 u ^{4}}{2 - u} d u

Take the constant out:

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

= 5 \cdot \int \frac{u ^{4}}{2 - u} d u

Long division

\frac{u ^{4}}{2 - u} : - u^{3} - 2 u^{2} - 4 u - 8 + \frac{16}{- u + 2}

= 5 \cdot \int - u^{3} - 2 u^{2} - 4 u - 8 + \frac{16}{- u + 2} 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

= 5 (- \int u^{3} d u - \int 2 u^{2} d u - \int 4 u d u - \int 8 d u + \int \frac{16}{- u + 2} d u)

\int u^{3} d u = \frac{u ^{4}}{4}

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

\int 4 u d u = 2 u^{2}

\int 8 d u = 8 u

\int \frac{16}{- u + 2} d u = - 16 ln ∣ - u + 2 ∣

= 5 (- \frac{u ^{4}}{4} - \frac{2 u ^{3}}{3} - 2 u^{2} - 8 u - 16 ln ∣ - u + 2 ∣)

Substitute back

u = x^{\frac{1}{5}}

= 5 - \frac{( x ^{\frac{1}{5}} ) ^{4}}{4} - \frac{2 ( x ^{\frac{1}{5}} ) ^{3}}{3} - 2 (x^{\frac{1}{5}})^{2} - 8 x^{\frac{1}{5}} - 16 ln - x^{\frac{1}{5}} + 2

Simplify

5 - \frac{( x ^{\frac{1}{5}} ) ^{4}}{4} - \frac{2 ( x ^{\frac{1}{5}} ) ^{3}}{3} - 2 (x^{\frac{1}{5}})^{2} - 8 x^{\frac{1}{5}} - 16 ln - x^{\frac{1}{5}} + 2 : 5 (- \frac{1}{4} x^{\frac{4}{5}} - \frac{2}{3} x^{\frac{3}{5}} - 2 x^{\frac{2}{5}} - 8 x^{\frac{1}{5}} - 16 ln - x^{\frac{1}{5}} + 2)

= 5 (- \frac{1}{4} x^{\frac{4}{5}} - \frac{2}{3} x^{\frac{3}{5}} - 2 x^{\frac{2}{5}} - 8 x^{\frac{1}{5}} - 16 ln - x^{\frac{1}{5}} + 2)

Add a constant to the solution

= 5 (- \frac{1}{4} x^{\frac{4}{5}} - \frac{2}{3} x^{\frac{3}{5}} - 2 x^{\frac{2}{5}} - 8 x^{\frac{1}{5}} - 16 ln - x^{\frac{1}{5}} + 2) + C

f (x) = \frac{x + x e ^{x}}{x ^{2} - 5}

\frac{d}{d x} (3 1 + tan (x^{2}))

x (3 - x)^{3}

f (x) = x^{2} cos (\frac{1}{x})

y = \frac{x + 1}{x - 1}

What is the integral of 1/(2-x^{1/5)} ?
The integral of 1/(2-x^{1/5)} is 5(-1/4 x^{4/5}-2/3 x^{3/5}-2x^{2/5}-8x^{1/5}-16ln|-x^{1/5}+2|)+C