What is the integral of 1/(x+5sqrt(x)+6) ?

The integral of 1/(x+5sqrt(x)+6) is ln|4x+20sqrt(x)+24|+5ln|2sqrt(x)+6|-5ln|2sqrt(x)+4|+C

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integral of 1/(x+5sqrt(x)+6)

\int \frac{1}{x + 5 x + 6} d x

ln 4 x + 20 x + 24 + 5 ln 2 x + 6 - 5 ln 2 x + 4 + C

Solution steps

\int \frac{1}{x + 5 x + 6} d x

Apply u-substitution

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

Complete the square

u^{2} + 5 u + 6 : (u + \frac{5}{2})^{2} - \frac{1}{4}

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

Apply u-substitution

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

Expand

\frac{2 v - 5}{4 v ^{2} - 1} : \frac{2 v}{4 v ^{2} - 1} - \frac{5}{4 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 \cdot 2 (\int \frac{2 v}{4 v ^{2} - 1} d v - \int \frac{5}{4 v ^{2} - 1} d v)

\int \frac{2 v}{4 v ^{2} - 1} d v = \frac{1}{4} ln 4 v^{2} - 1

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

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

Substitute back

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

Simplify

2 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{5}{2})^{2} - 1 - (- \frac{5}{4} (ln 2 (x + \frac{5}{2}) + 1 - ln 2 (x + \frac{5}{2}) - 1))) : ln 4 x + 20 x + 24 + 5 ln 2 x + 6 - 5 ln 2 x + 4

= ln 4 x + 20 x + 24 + 5 ln 2 x + 6 - 5 ln 2 x + 4

Add a constant to the solution

= ln 4 x + 20 x + 24 + 5 ln 2 x + 6 - 5 ln 2 x + 4 + C

What is the integral of 1/(x+5sqrt(x)+6) ?
The integral of 1/(x+5sqrt(x)+6) is ln|4x+20sqrt(x)+24|+5ln|2sqrt(x)+6|-5ln|2sqrt(x)+4|+C