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

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

Apply the Power Rule

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

= - \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

y^{^{'}} = - \frac{2}{x} y

\frac{x ^{2}}{x - 1}

\frac{d}{d x} (sin^{e^{x}} (x))

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

\int (x + 1)^{2} (1 - x)^{5} d x

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