What is the integral of ()/(t^3sqrt(t^2-9)) ?

The integral of ()/(t^3sqrt(t^2-9)) is (dt)/(sqrt(-9+t^2)t^3)x+C

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integral of ()/(t^3sqrt(t^2-9))

\int \frac{d t}{t ^{3} t ^{2} - 9} d x

\frac{d t}{- 9 + t ^{2} t ^{3}} x + C

Solution steps

\int \frac{d t}{t ^{3} t ^{2} - 9} d x

Integral of a constant:

\int a d x = a x

= \frac{d t}{- 9 + t ^{2} t ^{3}} x

Add a constant to the solution

= \frac{d t}{- 9 + t ^{2} t ^{3}} x + C

\frac{\partial}{\partial x} ((2 x + 5 y) e^{y})

\frac{\partial}{\partial x} (x + y - x y)

\int_{0}^{8} (x + 1)^{2} d x

y = arcsin (\frac{1}{x ^{5}})

x \to 0 lim (\frac{ln ( 1 - 6 x )}{x})

What is the integral of ()/(t^3sqrt(t^2-9)) ?
The integral of ()/(t^3sqrt(t^2-9)) is (dt)/(sqrt(-9+t^2)t^3)x+C