Not seeing how to integrate y/(y+1)

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SUMMARY

The integral of the function \(\frac{y}{y + 1}\) can be evaluated using algebraic manipulation and substitution techniques. By rewriting the integrand as \(1 - \frac{1}{y + 1}\), the integral simplifies to \(\int 1 \, dy - \int \frac{1}{y + 1} \, dy\), resulting in the solution \(y - \ln |y + 1| + C\). Alternatively, a substitution \(u = y + 1\) can also be employed, leading to the integral of \(\frac{u - 1}{u}\) with respect to \(u\).

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Hello,

I cannot figure out how to evaluate the following integral,

\int {\frac{y}{{y + 1}}dy}

If it was y^2+1 then I see how a u-substitution of u=y^2+1 would work.

Thanks
 
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Usually with fractions like this, you'll find you can express what you have as the sum of two or more fractions with the same denominator which are easier to integrate. In this case, try the numerators y+1 and -1.
 
As in, write y = y + 1 - 1.
 
Ok, that is what I was having trouble seeing. Thanks

<br /> \frac{y}{{y + 1}} = \frac{{y + 1}}{{y + 1}} - \frac{1}{{y + 1}} = 1 - \frac{1}{{y + 1}}<br />

<br /> \int {1{\rm }dy - } \int {\frac{1}{{y + 1}}{\rm }} dy = y - \ln \left| {y + 1} \right| + C<br />
 
You could also substitute u = y + 1, du = dy, and get the integral of (u - 1) / u with respect to u.
 

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