What is the integral of 1/(4xsqrt(16x^2-25)) ?

The integral of 1/(4xsqrt(16x^2-25)) is 1/20 arcsec(4/5 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4xsqrt(16x^2-25))

\int \frac{1}{4 x 16 x ^{2} - 25} d x

\frac{1}{20} arcsec (\frac{4}{5} x) + C

Solution steps

\int \frac{1}{4 x 16 x ^{2} - 25} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{x 16 x ^{2} - 25} d x

Apply Trigonometric Substitution

: \int \frac{1}{5} d u

= \frac{1}{4} \cdot \int \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{4} \cdot \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot \frac{1}{5} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{4}{5} x)

= \frac{1}{4} \cdot \frac{1}{5} \cdot 1 \cdot arcsec (\frac{4}{5} x)

Simplify

\frac{1}{4} \cdot \frac{1}{5} \cdot 1 \cdot arcsec (\frac{4}{5} x) : \frac{1}{20} arcsec (\frac{4}{5} x)

= \frac{1}{20} arcsec (\frac{4}{5} x)

Add a constant to the solution

= \frac{1}{20} arcsec (\frac{4}{5} x) + C

- 3 e^{- 3 x} + e^{4 x} \cdot 4

\int y^{2} e^{3 y} d y

n = 0 \sum \infty \frac{1}{n + 7}

\int_{0}^{9} \frac{8}{( x - 1 ) ^{\frac{1}{3}}} d x

\frac{\partial}{\partial x} (9 x \cdot cos (x) \cdot cos (y))

What is the integral of 1/(4xsqrt(16x^2-25)) ?
The integral of 1/(4xsqrt(16x^2-25)) is 1/20 arcsec(4/5 x)+C