What is the integral of 8/(y^4-16) ?

The integral of 8/(y^4-16) is 8(-1/16 arctan(y/2)-1/32 ln|y+2|+1/32 ln|y-2|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/(y^4-16)

\int \frac{8}{y ^{4} - 16} d y

8 (- \frac{1}{16} arctan (\frac{y}{2}) - \frac{1}{32} ln ∣ y + 2 ∣ + \frac{1}{32} ln ∣ y - 2 ∣) + C

Solution steps

\int \frac{8}{y ^{4} - 16} d y

Take the constant out:

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

= 8 \cdot \int \frac{1}{y ^{4} - 16} d y

Take the partial fraction of

\frac{1}{y ^{4} - 16} : - \frac{1}{8 ( y ^{2} + 4 )} - \frac{1}{32 ( y + 2 )} + \frac{1}{32 ( y - 2 )}

= 8 \cdot \int - \frac{1}{8 ( y ^{2} + 4 )} - \frac{1}{32 ( y + 2 )} + \frac{1}{32 ( y - 2 )} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (- \int \frac{1}{8 ( y ^{2} + 4 )} d y - \int \frac{1}{32 ( y + 2 )} d y + \int \frac{1}{32 ( y - 2 )} d y)

\int \frac{1}{8 ( y ^{2} + 4 )} d y = \frac{1}{16} arctan (\frac{y}{2})

\int \frac{1}{32 ( y + 2 )} d y = \frac{1}{32} ln ∣ y + 2 ∣

\int \frac{1}{32 ( y - 2 )} d y = \frac{1}{32} ln ∣ y - 2 ∣

= 8 (- \frac{1}{16} arctan (\frac{y}{2}) - \frac{1}{32} ln ∣ y + 2 ∣ + \frac{1}{32} ln ∣ y - 2 ∣)

Add a constant to the solution

= 8 (- \frac{1}{16} arctan (\frac{y}{2}) - \frac{1}{32} ln ∣ y + 2 ∣ + \frac{1}{32} ln ∣ y - 2 ∣) + C

\int - 5 e^{c o s (5 x)} sin (5 x) d x

\int sec (x) (7 sec (x) + 3 tan (x)) d x

\int (1 + t^{2}) d t

\int ln (x 1 + x^{2}) d x

\int cos (53 x) d x

What is the integral of 8/(y^4-16) ?
The integral of 8/(y^4-16) is 8(-1/16 arctan(y/2)-1/32 ln|y+2|+1/32 ln|y-2|)+C