Integral of 1(y1)2\frac{1}{\left(y - 1\right)^{2}}

The calculator will find the integral/antiderivative of 1(y1)2\frac{1}{\left(y - 1\right)^{2}}, with steps shown.

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Your Input

Find 1(y1)2dy\int \frac{1}{\left(y - 1\right)^{2}}\, dy.

Solution

Let u=y1u=y - 1.

Then du=(y1)dy=1dydu=\left(y - 1\right)^{\prime }dy = 1 dy (steps can be seen »), and we have that dy=dudy = du.

The integral becomes

1(y1)2dy=1u2du{\color{red}{\int{\frac{1}{\left(y - 1\right)^{2}} d y}}} = {\color{red}{\int{\frac{1}{u^{2}} d u}}}

Apply the power rule undu=un+1n+1\int u^{n}\, du = \frac{u^{n + 1}}{n + 1} (n1)\left(n \neq -1 \right) with n=2n=-2:

1u2du=u2du=u2+12+1=(u1)=(1u){\color{red}{\int{\frac{1}{u^{2}} d u}}}={\color{red}{\int{u^{-2} d u}}}={\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}={\color{red}{\left(- u^{-1}\right)}}={\color{red}{\left(- \frac{1}{u}\right)}}

Recall that u=y1u=y - 1:

u1=(y1)1- {\color{red}{u}}^{-1} = - {\color{red}{\left(y - 1\right)}}^{-1}

Therefore,

1(y1)2dy=1y1\int{\frac{1}{\left(y - 1\right)^{2}} d y} = - \frac{1}{y - 1}

Add the constant of integration:

1(y1)2dy=1y1+C\int{\frac{1}{\left(y - 1\right)^{2}} d y} = - \frac{1}{y - 1}+C

Answer

1(y1)2dy=1y1+C\int \frac{1}{\left(y - 1\right)^{2}}\, dy = - \frac{1}{y - 1} + CA