What is the integral of 1/(xsqrt(x-4)) ?

The integral of 1/(xsqrt(x-4)) is arctan(1/2 sqrt(x-4))+C

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integral of 1/(xsqrt(x-4))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 4} d u

= \int \frac{2}{u ^{2} + 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{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Use the common integral:

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

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} arctan (\frac{x - 4}{2})

Simplify

2 \cdot \frac{1}{2} arctan (\frac{x - 4}{2}) : arctan (\frac{1}{2} x - 4)

= arctan (\frac{1}{2} x - 4)

Add a constant to the solution

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

\int sec^{4} (\frac{x}{3}) d x

y = x^{2} e^{x} - 2 x e^{x} + 5 e^{x}

\int_{- π}^{π} cos^{2} (x) d x

\frac{d}{d x} (\frac{4}{x ^{2} - 1})

y = x^{2} + 2 x + 6

What is the integral of 1/(xsqrt(x-4)) ?
The integral of 1/(xsqrt(x-4)) is arctan(1/2 sqrt(x-4))+C