What is the integral of 1/(x+6+6sqrt(x+1)) ?

The integral of 1/(x+6+6sqrt(x+1)) is 2(5/4 ln|sqrt(x+1)+5|-1/4 ln|sqrt(x+1)+1|)+C

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integral of 1/(x+6+6sqrt(x+1))

\int \frac{1}{x + 6 + 6 x + 1} d x

2 (\frac{5}{4} ln x + 1 + 5 - \frac{1}{4} ln x + 1 + 1) + C

Solution steps

\int \frac{1}{x + 6 + 6 x + 1} d x

Apply u-substitution

: \int \frac{2 u}{u ^{2} + 6 u + 5} d u

= \int \frac{2 u}{u ^{2} + 6 u + 5} 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} + 6 u + 5} d u

Complete the square

u^{2} + 6 u + 5 : (u + 3)^{2} - 4

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

Apply u-substitution

: \int \frac{v - 3}{v ^{2} - 4} d v

= 2 \cdot \int \frac{v - 3}{v ^{2} - 4} d v

Take the partial fraction of

\frac{v - 3}{v ^{2} - 4} : \frac{5}{4 ( v + 2 )} - \frac{1}{4 ( v - 2 )}

= 2 \cdot \int \frac{5}{4 ( v + 2 )} - \frac{1}{4 ( v - 2 )} 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 \frac{5}{4 ( v + 2 )} d v - \int \frac{1}{4 ( v - 2 )} d v)

\int \frac{5}{4 ( v + 2 )} d v = \frac{5}{4} ln ∣ v + 2 ∣

\int \frac{1}{4 ( v - 2 )} d v = \frac{1}{4} ln ∣ v - 2 ∣

= 2 (\frac{5}{4} ln ∣ v + 2 ∣ - \frac{1}{4} ln ∣ v - 2 ∣)

Substitute back

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

Simplify

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

= 2 (\frac{5}{4} ln x + 1 + 5 - \frac{1}{4} ln x + 1 + 1)

Add a constant to the solution

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

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What is the integral of 1/(x+6+6sqrt(x+1)) ?
The integral of 1/(x+6+6sqrt(x+1)) is 2(5/4 ln|sqrt(x+1)+5|-1/4 ln|sqrt(x+1)+1|)+C