What is the integral of (sqrt(9x^2+16))/(x^4) ?

The integral of (sqrt(9x^2+16))/(x^4) is -((16+9x^2)^{3/2})/(48x^3)+C

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integral of (sqrt(9x^2+16))/(x^4)

\int \frac{9 x ^{2} + 16}{x ^{4}} d x

- \frac{( 16 + 9 x ^{2} ) ^{\frac{3}{2}}}{48 x ^{3}} + C

Solution steps

\int \frac{9 x ^{2} + 16}{x ^{4}} d x

Apply Trigonometric Substitution

: \int \frac{27 sec ^{3} ( u )}{16 tan ^{4} ( u )} d u

= \int \frac{27 sec ^{3} ( u )}{16 tan ^{4} ( u )} d u

Take the constant out:

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

= \frac{27}{16} \cdot \int \frac{sec ^{3} ( u )}{tan ^{4} ( u )} d u

Rewrite using trig identities

= \frac{27}{16} \cdot \int \frac{cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

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

= \frac{27}{16} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

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

= \frac{27}{16} (- \frac{1}{3 v ^{3}})

Substitute back

= \frac{27}{16} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{3}{4} x ) )})

Simplify

\frac{27}{16} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{3}{4} x ) )}) : - \frac{( 16 + 9 x ^{2} ) ^{\frac{3}{2}}}{48 x ^{3}}

= - \frac{( 16 + 9 x ^{2} ) ^{\frac{3}{2}}}{48 x ^{3}}

Add a constant to the solution

= - \frac{( 16 + 9 x ^{2} ) ^{\frac{3}{2}}}{48 x ^{3}} + C

\int 2 x (x^{2} - 1) d x

y = (\frac{x ^{2} + 3}{x ^{2} - 3})^{7}

x \to 2 lim (x^{2} + 5 x - 9)

y = 3 tan (3 t)

cot^{2} (cos (θ))

What is the integral of (sqrt(9x^2+16))/(x^4) ?
The integral of (sqrt(9x^2+16))/(x^4) is -((16+9x^2)^{3/2})/(48x^3)+C