What is the integral of (3y^2-t^2)/(y^5) ?

The integral of (3y^2-t^2)/(y^5) is (t^2)/(4y^4)-3/(2y^2)+C

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integral of (3y^2-t^2)/(y^5)

\int \frac{3 y ^{2} - t ^{2}}{y ^{5}} d y

\frac{t ^{2}}{4 y ^{4}} - \frac{3}{2 y ^{2}} + C

Solution steps

\int \frac{3 y ^{2} - t ^{2}}{y ^{5}} d y

Expand

\frac{- t ^{2} + 3 y ^{2}}{y ^{5}} : - \frac{t ^{2}}{y ^{5}} + \frac{3}{y ^{3}}

= \int - \frac{t ^{2}}{y ^{5}} + \frac{3}{y ^{3}} d y

Apply the Sum Rule:

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

= - \int \frac{t ^{2}}{y ^{5}} d y + \int \frac{3}{y ^{3}} d y

\int \frac{t ^{2}}{y ^{5}} d y = - \frac{t ^{2}}{4 y ^{4}}

\int \frac{3}{y ^{3}} d y = - \frac{3}{2 y ^{2}}

= - (- \frac{t ^{2}}{4 y ^{4}}) - \frac{3}{2 y ^{2}}

Simplify

= \frac{t ^{2}}{4 y ^{4}} - \frac{3}{2 y ^{2}}

Add a constant to the solution

= \frac{t ^{2}}{4 y ^{4}} - \frac{3}{2 y ^{2}} + C

\frac{d}{d x} (ln (9) x^{2})

x \to - 8 - lim (\frac{- 5 x}{x + 8})

\int_{3}^{9} (1 + x^{2}) d x

\int \frac{24}{( x - 2 ) ( x ^{2} + 4 )} d x

\int_{0}^{1} \frac{x ^{2}}{x ^{3} - 1} d x

What is the integral of (3y^2-t^2)/(y^5) ?
The integral of (3y^2-t^2)/(y^5) is (t^2)/(4y^4)-3/(2y^2)+C