What is the integral of 1/((a+by)^3) ?

The integral of 1/((a+by)^3) is y/((a+by)^3)-3/(2b(by+a)^2)+a/(b(by+a)^3)+C

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integral of 1/((a+by)^3)

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

\frac{y}{( a + b y ) ^{3}} - \frac{3}{2 b ( b y + a ) ^{2}} + \frac{a}{b ( b y + a ) ^{3}} + C

Solution steps

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

Apply Integration By Parts

: \frac{y}{( a + b y ) ^{3}} - \int - \frac{3 b y}{( a + b y ) ^{4}} d y

= \frac{y}{( a + b y ) ^{3}} - \int - \frac{3 b y}{( a + b y ) ^{4}} d y

\int - \frac{3 b y}{( a + b y ) ^{4}} d y = \frac{3}{2 b ( b y + a ) ^{2}} - \frac{a}{b ( b y + a ) ^{3}}

= \frac{y}{( a + b y ) ^{3}} - (\frac{3}{2 b ( b y + a ) ^{2}} - \frac{a}{b ( b y + a ) ^{3}})

Simplify

= \frac{y}{( a + b y ) ^{3}} - \frac{3}{2 b ( b y + a ) ^{2}} + \frac{a}{b ( b y + a ) ^{3}}

Add a constant to the solution

= \frac{y}{( a + b y ) ^{3}} - \frac{3}{2 b ( b y + a ) ^{2}} + \frac{a}{b ( b y + a ) ^{3}} + C

\int \frac{y}{4 - y ^{2}} d y

\frac{\partial}{\partial z} (5 x^{2} - 3 x y + x yz)

\frac{\partial}{\partial t} (ln (3 x + 3 c t))

x \to 0 lim (\frac{1}{x} - \frac{1}{e ^{x}})

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 16 )}

What is the integral of 1/((a+by)^3) ?
The integral of 1/((a+by)^3) is y/((a+by)^3)-3/(2b(by+a)^2)+a/(b(by+a)^3)+C