What is the integral of 1/(y^6)-9/(sqrt(y)) ?

The integral of 1/(y^6)-9/(sqrt(y)) is -1/(5y^5)-18sqrt(y)+C

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integral of 1/(y^6)-9/(sqrt(y))

\int \frac{1}{y ^{6}} - \frac{9}{y} d y

- \frac{1}{5 y ^{5}} - 18 y + C

Solution steps

\int \frac{1}{y ^{6}} - \frac{9}{y} 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 ^{6}} d y - \int \frac{9}{y} d y

\int \frac{1}{y ^{6}} d y = - \frac{1}{5 y ^{5}}

\int \frac{9}{y} d y = 18 y

= - \frac{1}{5 y ^{5}} - 18 y

Add a constant to the solution

= - \frac{1}{5 y ^{5}} - 18 y + C

(4 a + 1)^{2}

y^{^{''}} + 4 y^{^{'}} = 10 e^{t}

\frac{d}{d x} (\frac{e ^{4 x} + e ^{- 4 x}}{x})

2 x, x = 3

f^{^{''}} (x) = f (x)

What is the integral of 1/(y^6)-9/(sqrt(y)) ?
The integral of 1/(y^6)-9/(sqrt(y)) is -1/(5y^5)-18sqrt(y)+C