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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

= 4 (- \frac{x ^{2} + 16}{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}

= 4 (- \frac{x ^{2} + 16}{3 x ^{3}} - \frac{16 + x ^{2}}{48 x})

Add a constant to the solution

= 4 (- \frac{x ^{2} + 16}{3 x ^{3}} - \frac{16 + x ^{2}}{48 x}) + C

\int - sec^{2} (\frac{7 x}{6}) d x

n = 4 \sum \infty \frac{6}{n ^{2} - 1}

x \to 0 lim (e^{- x (1 - s)})

x \to π lim (sin (x))

f (x) = 5 sin (x) - 3 cos (x), at x = π

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