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$\color{green}{question}$:

How do you evaluate this integral?

$$\int \frac{y^2}{y^2+d^2}dy=y-d\,{\tan}^{-1}\left ( \frac{y}{d} \right )+\mathrm{constant}$$

$\color{green}{I~know}$ I should use the change of variables, But I do not know how to do.

Thank you for any hint.

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3 Answers3

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$$\frac{y^2}{y^2+d^2} = \frac{y^2 + d^2}{y^2+d^2} - \frac{d^2}{y^2+d^2}$$

The latter is an arctangent integral.

Eric Auld
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6

Hint

  • $$\frac{y^2}{y^2+d^2}=\frac{y^2+d^2-d^2}{y^2+d^2}=1-\frac{d^2}{y^2+d^2}$$
  • $$\frac{d^2}{y^2+d^2}=\frac{1}{(y/d)^2+1}=\frac{1}{t^2+1}$$
  • $$\int\frac{dt}{t^2+1}=\arctan t+C$$
0

Maple code with(Student[Calculus1]):IntTutor$(y^2/(y^2 +d^2),y$); produced the output which can be seen here. It coincides with the excellent answer by Eric Auld.

Harish Kayarohanam
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user64494
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