What is the integral of 1/(4tsqrt(t^2-16)) ?

The integral of 1/(4tsqrt(t^2-16)) is 1/16 arcsec(1/4 t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4tsqrt(t^2-16))

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

\frac{1}{16} arcsec (\frac{1}{4} t) + C

Solution steps

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

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}{t t ^{2} - 16} d t

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

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

Integral of a constant:

\int a d x = a x

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

Substitute back

u = arcsec (\frac{1}{4} t)

= \frac{1}{4} \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} t)

Simplify

\frac{1}{4} \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} t) : \frac{1}{16} arcsec (\frac{1}{4} t)

= \frac{1}{16} arcsec (\frac{1}{4} t)

Add a constant to the solution

= \frac{1}{16} arcsec (\frac{1}{4} t) + C

y^{^{'}} = 2 x^{2}

e^{2 x} + x e^{2 x} + cos (2 x)

\frac{\partial}{\partial x} (ln (y cos (x)))

y = e^{x}, at x = ln (2)

\int x (2 x + 5)^{20} d x

What is the integral of 1/(4tsqrt(t^2-16)) ?
The integral of 1/(4tsqrt(t^2-16)) is 1/16 arcsec(1/4 t)+C