What is the integral of (sqrt(25x^2+4))/(x^4) ?

The integral of (sqrt(25x^2+4))/(x^4) is -((4+25x^2)^{3/2})/(12x^3)+C

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integral of (sqrt(25x^2+4))/(x^4)

\int \frac{25 x ^{2} + 4}{x ^{4}} d x

- \frac{( 4 + 25 x ^{2} ) ^{\frac{3}{2}}}{12 x ^{3}} + C

Solution steps

\int \frac{25 x ^{2} + 4}{x ^{4}} d x

Apply Trigonometric Substitution

: \int \frac{125 sec ^{3} ( u )}{4 tan ^{4} ( u )} d u

= \int \frac{125 sec ^{3} ( u )}{4 tan ^{4} ( u )} d u

Take the constant out:

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

= \frac{125}{4} \cdot \int \frac{sec ^{3} ( u )}{tan ^{4} ( u )} d u

Use the basic trigonometric identity:

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

= \frac{125}{4} \cdot \int cot^{4} (u) sec^{3} (u) d u

Rewrite using trig identities

= \frac{125}{4} \cdot \int \frac{cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

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

= \frac{125}{4} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

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

= \frac{125}{4} (- \frac{1}{3 v ^{3}})

Substitute back

= \frac{125}{4} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{5}{2} x ) )})

Simplify

\frac{125}{4} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{5}{2} x ) )}) : - \frac{( 4 + 25 x ^{2} ) ^{\frac{3}{2}}}{12 x ^{3}}

= - \frac{( 4 + 25 x ^{2} ) ^{\frac{3}{2}}}{12 x ^{3}}

Add a constant to the solution

= - \frac{( 4 + 25 x ^{2} ) ^{\frac{3}{2}}}{12 x ^{3}} + C

f (x) = ln (sin (2 x))

\frac{\partial}{\partial x} (ln (tan (x + y)))

\int_{1}^{- 1} (3 e^{x}) d x

\frac{d}{d x} (x e^{x^{3}})

\frac{1}{( s + 3 ) ^{2} + 4}

What is the integral of (sqrt(25x^2+4))/(x^4) ?
The integral of (sqrt(25x^2+4))/(x^4) is -((4+25x^2)^{3/2})/(12x^3)+C