What is the integral of 1/(x+4sqrt(x-3)) ?

The integral of 1/(x+4sqrt(x-3)) is ln|x+4sqrt(x-3)|+2ln|sqrt(x-3)+3|-2ln|sqrt(x-3)+1|+C

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integral of 1/(x+4sqrt(x-3))

\int \frac{1}{x + 4 x - 3} d x

ln x + 4 x - 3 + 2 ln x - 3 + 3 - 2 ln x - 3 + 1 + C

Solution steps

\int \frac{1}{x + 4 x - 3} d x

Apply u-substitution

: \int \frac{2 u}{u ^{2} + 3 + 4 u} d u

= \int \frac{2 u}{u ^{2} + 3 + 4 u} 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 ^{2} + 3 + 4 u} d u

Complete the square

u^{2} + 3 + 4 u : (u + 2)^{2} - 1

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

Apply u-substitution

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

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

Expand

\frac{v - 2}{v ^{2} - 1} : \frac{v}{v ^{2} - 1} - \frac{2}{v ^{2} - 1}

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

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

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

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

Substitute back

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

Simplify

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

= ln x + 4 x - 3 + 2 ln x - 3 + 3 - 2 ln x - 3 + 1

Add a constant to the solution

= ln x + 4 x - 3 + 2 ln x - 3 + 3 - 2 ln x - 3 + 1 + C

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What is the integral of 1/(x+4sqrt(x-3)) ?
The integral of 1/(x+4sqrt(x-3)) is ln|x+4sqrt(x-3)|+2ln|sqrt(x-3)+3|-2ln|sqrt(x-3)+1|+C