What is the integral of 1/((3+y)^5) ?

The integral of 1/((3+y)^5) is -1/(4(3+y)^4)+C

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integral of 1/((3+y)^5)

\int \frac{1}{( 3 + y ) ^{5}} d y

- \frac{1}{4 ( 3 + y ) ^{4}} + C

Solution steps

\int \frac{1}{( 3 + y ) ^{5}} d y

Apply u-substitution

= \int \frac{1}{u ^{5}} d u

Apply the Power Rule

= - \frac{1}{4 u ^{4}}

Substitute back

u = 3 + y

= - \frac{1}{4 ( 3 + y ) ^{4}}

Add a constant to the solution

= - \frac{1}{4 ( 3 + y ) ^{4}} + C

s im pl i f y \frac{x ^{2}}{x - 1}

\int - \frac{1300}{x ^{2}} d x

\frac{d}{d x} (\frac{3 x}{x})

t an g e n t f (x) = (2)^{x}, at x = 6

y^{^{''}} + 14 y^{^{'}} + 28 y = 6675 e^{2 t} cos (15 t)

What is the integral of 1/((3+y)^5) ?
The integral of 1/((3+y)^5) is -1/(4(3+y)^4)+C