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So the Integral is $$\int \frac {dt}{t(t+1)^2}$$ So i thought of it two ways.

1.Substituting $\displaystyle y=\frac 1t$

Solution: $\displaystyle \ln\left|\frac{t}{1+t}\right|-\frac{t}{1+t}$

2.Substituting $\displaystyle y=\frac {1}{1+t}$

Solution: $\displaystyle \ln\left|\frac{t}{1+t}\right|+\frac {1}{1+t}$

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    $\begingroup$ Both look OK, they differ by a constant. +C... $\endgroup$ Commented Apr 17, 2023 at 12:25
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    $\begingroup$ Your answers differ by the constant $1$. So, both are valid. Add $1$ to your first answer and see. You'll get the second answer. $\endgroup$ Commented Apr 17, 2023 at 12:50
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    $\begingroup$ When in doubt that the integral is wrong, compute the derivative $\endgroup$ Commented Apr 17, 2023 at 12:55
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    $\begingroup$ Does this answer your question? Getting different answers when integrating using different techniques $\endgroup$ Commented Apr 17, 2023 at 14:13

1 Answer 1

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1.Substituting $\displaystyle y=\frac 1t$

Solution: $\displaystyle \ln\left|\frac{t}{1+t}\right|-\frac{t}{1+t}$

2.Substituting $\displaystyle y=\frac {1}{1+t}$

Solution: $\displaystyle \ln\left|\frac{t}{1+t}\right|+\frac {1}{1+t}$

Note that:

$$\frac{1}{1+t}=-\frac{t}{1+t}+1$$

your two solutions differ by a constant $1$, hence both of them are correct.

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