What is the integral of 1/(sqrt(16x^2-24x-27)) ?

The integral of 1/(sqrt(16x^2-24x-27)) is 1/4 ln((|sqrt(16x^2-24x-27)+4x-3|)/6)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(16x^2-24x-27))

\int \frac{1}{16 x ^{2} - 24 x - 27} d x

\frac{1}{4} ln (\frac{16 x ^{2} - 24 x - 27 + 4 x - 3}{6}) + C

Solution steps

\int \frac{1}{16 x ^{2} - 24 x - 27} d x

Complete the square

16 x^{2} - 24 x - 27 : 16 (x - \frac{3}{4})^{2} - 36

= \int \frac{1}{16 ( x - \frac{3}{4} ) ^{2} - 36} d x

Apply u-substitution

: \int \frac{1}{2 4 u ^{2} - 9} d u

= \int \frac{1}{2 4 u ^{2} - 9} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{4 u ^{2} - 9} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= \frac{1}{2} \cdot \int \frac{sec ( v )}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} ln tan (arcsec (\frac{2}{3} (x - \frac{3}{4}))) + sec (arcsec (\frac{2}{3} (x - \frac{3}{4})))

Simplify

\frac{1}{2} \cdot \frac{1}{2} ln tan (arcsec (\frac{2}{3} (x - \frac{3}{4}))) + sec (arcsec (\frac{2}{3} (x - \frac{3}{4}))) : \frac{1}{4} ln (\frac{16 x ^{2} - 24 x - 27 + 4 x - 3}{6})

= \frac{1}{4} ln (\frac{16 x ^{2} - 24 x - 27 + 4 x - 3}{6})

Add a constant to the solution

= \frac{1}{4} ln (\frac{16 x ^{2} - 24 x - 27 + 4 x - 3}{6}) + C

\frac{x}{5}

y = 7 cos (x)

\int t^{2} \cdot ln^{2} (t) d t

x \to 0 lim (x^{- 3})

\frac{d}{d x} (4 x^{- \frac{1}{3}} + \frac{9}{x ^{\frac{5}{3}}})

What is the integral of 1/(sqrt(16x^2-24x-27)) ?
The integral of 1/(sqrt(16x^2-24x-27)) is 1/4 ln((|sqrt(16x^2-24x-27)+4x-3|)/6)+C