What is the integral of 1/(t^4sqrt(9+t^2)) ?

The integral of 1/(t^4sqrt(9+t^2)) is 1/81 (-((9+t^2)^{3/2}}{3t^3}+\frac{sqrt(9+t^2))/t)+C

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integral of 1/(t^4sqrt(9+t^2))

\int \frac{1}{t ^{4} 9 + t ^{2}} d t

\frac{1}{81} - \frac{( 9 + t ^{2} ) ^{\frac{3}{2}}}{3 t ^{3}} + \frac{9 + t ^{2}}{t} + C

Solution steps

\int \frac{1}{t ^{4} 9 + t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{81 tan ^{4} ( u )} d u

= \int \frac{sec ( u )}{81 tan ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int \frac{sec ( u )}{tan ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{81} \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

: \int \frac{1 - v ^{2}}{v ^{4}} d v

= \frac{1}{81} \cdot \int \frac{1 - v ^{2}}{v ^{4}} d v

Expand

\frac{1 - v ^{2}}{v ^{4}} : \frac{1}{v ^{4}} - \frac{1}{v ^{2}}

= \frac{1}{81} \cdot \int \frac{1}{v ^{4}} - \frac{1}{v ^{2}} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{81} (\int \frac{1}{v ^{4}} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

= \frac{1}{81} (- \frac{1}{3 v ^{3}} - (- \frac{1}{v}))

Substitute back

= \frac{1}{81} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{3} t ) )} - (- \frac{1}{sin ( arctan ( \frac{1}{3} t ) )}))

Simplify

\frac{1}{81} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{3} t ) )} - (- \frac{1}{sin ( arctan ( \frac{1}{3} t ) )})) : \frac{1}{81} - \frac{( 9 + t ^{2} ) ^{\frac{3}{2}}}{3 t ^{3}} + \frac{9 + t ^{2}}{t}

= \frac{1}{81} - \frac{( 9 + t ^{2} ) ^{\frac{3}{2}}}{3 t ^{3}} + \frac{9 + t ^{2}}{t}

Add a constant to the solution

= \frac{1}{81} - \frac{( 9 + t ^{2} ) ^{\frac{3}{2}}}{3 t ^{3}} + \frac{9 + t ^{2}}{t} + C

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What is the integral of 1/(t^4sqrt(9+t^2)) ?
The integral of 1/(t^4sqrt(9+t^2)) is 1/81 (-((9+t^2)^{3/2}}{3t^3}+\frac{sqrt(9+t^2))/t)+C