What is the integral of (sqrt(4x^2+9))/(x^4) ?

The integral of (sqrt(4x^2+9))/(x^4) is -((9+4x^2)^{3/2})/(27x^3)+C

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integral of (sqrt(4x^2+9))/(x^4)

\int \frac{4 x ^{2} + 9}{x ^{4}} d x

- \frac{( 9 + 4 x ^{2} ) ^{\frac{3}{2}}}{27 x ^{3}} + C

Solution steps

\int \frac{4 x ^{2} + 9}{x ^{4}} d x

Apply Trigonometric Substitution

: \int \frac{8 sec ^{3} ( u )}{9 tan ^{4} ( u )} d u

= \int \frac{8 sec ^{3} ( u )}{9 tan ^{4} ( u )} d u

Take the constant out:

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

= \frac{8}{9} \cdot \int \frac{sec ^{3} ( u )}{tan ^{4} ( u )} d u

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

= \frac{8}{9} \cdot \int cot^{4} (u) sec^{3} (u) d u

Rewrite using trig identities

= \frac{8}{9} \cdot \int \frac{cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

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

= \frac{8}{9} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

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

= \frac{8}{9} (- \frac{1}{3 v ^{3}})

Substitute back

= \frac{8}{9} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{2}{3} x ) )})

Simplify

\frac{8}{9} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{2}{3} x ) )}) : - \frac{( 9 + 4 x ^{2} ) ^{\frac{3}{2}}}{27 x ^{3}}

= - \frac{( 9 + 4 x ^{2} ) ^{\frac{3}{2}}}{27 x ^{3}}

Add a constant to the solution

= - \frac{( 9 + 4 x ^{2} ) ^{\frac{3}{2}}}{27 x ^{3}} + C

\int \frac{x}{9 + x ^{4}} d x

\frac{1 + 0.5 x}{1 - 0.5 x}

\frac{d}{d x} (\frac{3 x}{2 x ^{2} + 1})

\frac{\partial}{\partial x} (\frac{1}{2} (x^{2} + 6 y^{2}))

f (x) = x (x^{2} - 4 x + 5)^{4}, at x = 2

What is the integral of (sqrt(4x^2+9))/(x^4) ?
The integral of (sqrt(4x^2+9))/(x^4) is -((9+4x^2)^{3/2})/(27x^3)+C