What is the integral of y^2(1/(y^5)+(y^4)/2) ?

The integral of y^2(1/(y^5)+(y^4)/2) is -1/(2y^2)+(y^7)/(14)+C

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integral of y^2(1/(y^5)+(y^4)/2)

\int y^{2} (\frac{1}{y ^{5}} + \frac{y ^{4}}{2}) d y

- \frac{1}{2 y ^{2}} + \frac{y ^{7}}{14} + C

Solution steps

\int y^{2} (\frac{1}{y ^{5}} + \frac{y ^{4}}{2}) d y

Expand

y^{2} (\frac{1}{y ^{5}} + \frac{y ^{4}}{2}) : \frac{1}{y ^{3}} + \frac{y ^{6}}{2}

= \int \frac{1}{y ^{3}} + \frac{y ^{6}}{2} 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{1}{y ^{3}} d y + \int \frac{y ^{6}}{2} d y

\int \frac{1}{y ^{3}} d y = - \frac{1}{2 y ^{2}}

\int \frac{y ^{6}}{2} d y = \frac{y ^{7}}{14}

= - \frac{1}{2 y ^{2}} + \frac{y ^{7}}{14}

Add a constant to the solution

= - \frac{1}{2 y ^{2}} + \frac{y ^{7}}{14} + C

6 x arcsin (x)

(t + e^{t}) (9 - t)

3 y^{^{'''}} + 12 y^{^{''}} + 12 y^{^{'}} = 0

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

\int_{0}^{1} e^{- t^{2}} d t

What is the integral of y^2(1/(y^5)+(y^4)/2) ?
The integral of y^2(1/(y^5)+(y^4)/2) is -1/(2y^2)+(y^7)/(14)+C