What is the integral of 1/(x^{1/2)-4} ?

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

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integral of 1/(x^{1/2)-4}

\int \frac{1}{x ^{\frac{1}{2}} - 4} d x

2 (x^{\frac{1}{2}} - 4 + 4 ln x^{\frac{1}{2}} - 4) + C

Solution steps

\int \frac{1}{x ^{\frac{1}{2}} - 4} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v + 4}{v} d v

= 2 \cdot \int \frac{v + 4}{v} d v

Expand

\frac{v + 4}{v} : 1 + \frac{4}{v}

= 2 \cdot \int 1 + \frac{4}{v} d v

Apply the Sum Rule:

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

= 2 (\int 1 d v + \int \frac{4}{v} d v)

\int 1 d v = v

\int \frac{4}{v} d v = 4 ln ∣ v ∣

= 2 (v + 4 ln ∣ v ∣)

Substitute back

= 2 (x^{\frac{1}{2}} - 4 + 4 ln x^{\frac{1}{2}} - 4)

Add a constant to the solution

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

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\frac{\partial}{\partial y} (7 e^{x} ln (y))

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