What is the integral of (sqrt(16x^2+25))/(x^4) ?

The integral of (sqrt(16x^2+25))/(x^4) is -((25+16x^2)^{3/2})/(75x^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(16x^2+25))/(x^4)

\int \frac{16 x ^{2} + 25}{x ^{4}} d x

- \frac{( 25 + 16 x ^{2} ) ^{\frac{3}{2}}}{75 x ^{3}} + C

Solution steps

\int \frac{16 x ^{2} + 25}{x ^{4}} d x

Apply Trigonometric Substitution

: \int \frac{64 sec ^{3} ( u )}{25 tan ^{4} ( u )} d u

= \int \frac{64 sec ^{3} ( u )}{25 tan ^{4} ( u )} d u

Take the constant out:

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

= \frac{64}{25} \cdot \int \frac{sec ^{3} ( u )}{tan ^{4} ( u )} d u

Use the basic trigonometric identity:

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

= \frac{64}{25} \cdot \int cot^{4} (u) sec^{3} (u) d u

Rewrite using trig identities

= \frac{64}{25} \cdot \int \frac{cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

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

= \frac{64}{25} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

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

= \frac{64}{25} (- \frac{1}{3 v ^{3}})

Substitute back

= \frac{64}{25} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{4}{5} x ) )})

Simplify

\frac{64}{25} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{4}{5} x ) )}) : - \frac{( 25 + 16 x ^{2} ) ^{\frac{3}{2}}}{75 x ^{3}}

= - \frac{( 25 + 16 x ^{2} ) ^{\frac{3}{2}}}{75 x ^{3}}

Add a constant to the solution

= - \frac{( 25 + 16 x ^{2} ) ^{\frac{3}{2}}}{75 x ^{3}} + C

\int_{0}^{π} cos (2 θ) d θ

\int_{0}^{2} 4 x^{3} - 3 x^{2} + 2 x d x

\frac{( 3 s + 5 )}{( s - 5 ) ( s + 2 )}

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

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

What is the integral of (sqrt(16x^2+25))/(x^4) ?
The integral of (sqrt(16x^2+25))/(x^4) is -((25+16x^2)^{3/2})/(75x^3)+C