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Evaluation of Integration $\displaystyle \int \frac{4}{x+9}dx$ without using $u$ substitution.

What i try

$$4\int\frac{(x+9)-x}{x+9}dx=4\int dx-4\int\frac{x}{x+9}dx$$

How do I solve it without using $u$ substitution . Help me please.

I did not understand how one can able to solve without using $u$ substitution.

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    $\begingroup$ Are you allowed the "notice" that $\frac{d}{dx}\log(x+9)=\frac{1}{x+9}$? $\endgroup$ – Reveillark Apr 17 '20 at 22:32
  • $\begingroup$ Frankly, it's just silly not to use substitution. Without substitution of some sort, you can only evaluate a small handful of antiderivatives. For example, you couldn't evaluate $\int(1+x)^{1000}\;dx$ $\endgroup$ – MPW Apr 17 '20 at 22:37
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    $\begingroup$ What is a $u$ substitution, as opposed to a good old substitution? $\endgroup$ – Bernard Apr 17 '20 at 22:44
  • $\begingroup$ You may take two cases : $x <9$ or $x>9$. In the former case, the integrand in question will have the term $(1+x/9)^{-1}$. Write its Maclaurin series and then integrate. Similarly for the latter case, do the same for $(1+9/x)^{-1}$. $\endgroup$ – Koro Apr 17 '20 at 22:55
  • $\begingroup$ What you tried actually made this more complicated. $\endgroup$ – 1123581321 Apr 17 '20 at 23:36
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Try using: $$ \int{\frac{a\cdot f'(x)}{f(x)}} = a\cdot \ln|f(x)|+C $$ Very useful to be aware of that.

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  • $\begingroup$ OP doesn't want to use substitution, and to some degree I can foresee this being interpreted as such since its proof ultimately relies on it. $\endgroup$ – Eevee Trainer Apr 17 '20 at 22:53
  • $\begingroup$ @Eevee Trainer: Why, just differentiate RHS(?) $\endgroup$ – farud Apr 17 '20 at 23:02
  • $\begingroup$ I suppose that would work too. $\endgroup$ – Eevee Trainer Apr 17 '20 at 23:04
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Without using $u$-substitution, I can think of using that the integrand is a geometric progression in disguise. And further using the Taylor series of $\ln(1+x)$, we can get the antiderivative.

$$\int\frac{4/9}{1+x/9}\mathrm dx=\frac{4}{9}\int\sum_{i=0}^{\infty}\frac{(-1)^{i}}{9^{i}}x^{i}\mathrm dx=4\sum_{i=0}^{\infty}\frac{(-1)^{i}}{9^{i+1}}\frac{x^{i+1}}{i+1}=4\ln\left(1+\frac{x}{9}\right)+\text{const.}$$

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