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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(1+((4x^4-1)^2)/(16x^4))

\int 1 + \frac{( 4 x ^{4} - 1 ) ^{2}}{16 x ^{4}} d x

\frac{x ^{3}}{3} - \frac{1}{4 x} + C

Solution steps

\int 1 + \frac{( 4 x ^{4} - 1 ) ^{2}}{16 x ^{4}} d x

Simplify

1 + \frac{( 4 x ^{4} - 1 ) ^{2}}{16 x ^{4}} : \frac{( 2 x ^{2} + 2 x + 1 ) ( 2 x ^{2} - 2 x + 1 )}{4 x ^{2}}

= \int \frac{( 2 x ^{2} + 2 x + 1 ) ( 2 x ^{2} - 2 x + 1 )}{4 x ^{2}} 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{( 2 x ^{2} + 2 x + 1 ) ( 2 x ^{2} - 2 x + 1 )}{x ^{2}} d x

Expand

\frac{( 2 x ^{2} + 2 x + 1 ) ( 2 x ^{2} - 2 x + 1 )}{x ^{2}} : 4 x^{2} + \frac{1}{x ^{2}}

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

Apply the Sum Rule:

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

= \frac{1}{4} (\int 4 x^{2} d x + \int \frac{1}{x ^{2}} d x)

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= \frac{1}{4} (\frac{4 x ^{3}}{3} - \frac{1}{x})

Simplify

\frac{1}{4} (\frac{4 x ^{3}}{3} - \frac{1}{x}) : \frac{x ^{3}}{3} - \frac{1}{4 x}

= \frac{x ^{3}}{3} - \frac{1}{4 x}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{1}{4 x} + C

\int (x^{2} - x (4 x - 2)) d x

\int \frac{8 x}{2 x + 7} d x

\int \frac{2 e ^{x}}{e ^{3 x} + 2 e ^{x}} d x

\int (x^{- 1} + 3)^{3} d x

\int sec^{2} (a x) d x

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