# integration without using $u$ substitution

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.

• Are you allowed the "notice" that $\frac{d}{dx}\log(x+9)=\frac{1}{x+9}$? – Reveillark Apr 17 '20 at 22:32
• 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 $\int(1+x)^{1000}\;dx$ – MPW Apr 17 '20 at 22:37
• What is a $u$ substitution, as opposed to a good old substitution? – Bernard Apr 17 '20 at 22:44
• 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 $(1+9/x)^{-1}$. – Koro Apr 17 '20 at 22:55
• What you tried actually made this more complicated. – 1123581321 Apr 17 '20 at 23:36

Try using: $$\int{\frac{a\cdot f'(x)}{f(x)}} = a\cdot \ln|f(x)|+C$$ Very useful to be aware of that.
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.
$$\int\frac{4/9}{1+x/9}\mathrm dx=\frac{4}{9}\int\sum_{i=0}^{\infty}\frac{(-1)^{i}}{9^{i}}x^{i}\mathrm dx=4\sum_{i=0}^{\infty}\frac{(-1)^{i}}{9^{i+1}}\frac{x^{i+1}}{i+1}=4\ln\left(1+\frac{x}{9}\right)+\text{const.}$$