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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{( 1 + 4 x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + C

Solution steps

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

Apply Trigonometric Substitution

: \int 8 cot^{4} (u) sec^{3} (u) d u

= \int 8 cot^{4} (u) sec^{3} (u) d u

Take the constant out:

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

= 8 \cdot \int cot^{4} (u) sec^{3} (u) d u

Rewrite using trig identities

= 8 \cdot \int \frac{cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{4}} d v

= 8 \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

: - \frac{1}{3 v ^{3}}

= 8 (- \frac{1}{3 v ^{3}})

Substitute back

= 8 (- \frac{1}{3 sin ^{3} ( arctan ( 2 x ) )})

Simplify

8 (- \frac{1}{3 sin ^{3} ( arctan ( 2 x ) )}) : - \frac{( 1 + 4 x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}}

= - \frac{( 1 + 4 x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}}

Add a constant to the solution

= - \frac{( 1 + 4 x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + C

\int x cos (5 π x^{2}) d x

\int z e^{x} d x

\int [x \cdot e^{- x}] d x

\int 1 - 2 t d t

\int (x^{2} - 5 x) d x

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