What is the integral of (sqrt(16+x^2))/(2x^4) ?

The integral of (sqrt(16+x^2))/(2x^4) is 1/2 (-(sqrt(16+x^2))/(3x^3)-(sqrt(16+x^2))/(48x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(16+x^2))/(2x^4)

\int \frac{16 + x ^{2}}{2 x ^{4}} d x

\frac{1}{2} (- \frac{16 + x ^{2}}{3 x ^{3}} - \frac{16 + x ^{2}}{48 x}) + C

Solution steps

\int \frac{16 + x ^{2}}{2 x ^{4}} d x

Take the constant out:

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

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

Apply Integration By Parts

: - \frac{16 + x ^{2}}{3 x ^{3}} - \int - \frac{1}{3 x ^{2} x ^{2} + 16} d x

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

\int - \frac{1}{3 x ^{2} x ^{2} + 16} d x = \frac{16 + x ^{2}}{48 x}

= \frac{1}{2} (- \frac{16 + x ^{2}}{3 x ^{3}} - \frac{16 + x ^{2}}{48 x})

Add a constant to the solution

= \frac{1}{2} (- \frac{16 + x ^{2}}{3 x ^{3}} - \frac{16 + x ^{2}}{48 x}) + C

\int (2 x^{2} - 1)^{\frac{1}{2}} d x

f (x) = ln (cos (x))

\frac{d}{d x} (ln (x^{2} + 2))

n = 1 \sum \infty (x + 1)^{n}

x \to - \infty lim (- 18 x^{4})

What is the integral of (sqrt(16+x^2))/(2x^4) ?
The integral of (sqrt(16+x^2))/(2x^4) is 1/2 (-(sqrt(16+x^2))/(3x^3)-(sqrt(16+x^2))/(48x))+C