Quotient rule integration problem

[thereddevils]

$$\int \frac{1}{2x+3}=\frac{\ln |2x+3|}{2}+c$$

so why is $$\int \frac{1}{x^2+x}\neq \frac{\ln |x^2+x|}{2x+1}+c$$ ?

is it because in general ,

$$\int \frac{1}{x}=\ln |x|+c$$

the denominator is meant to be only linear function ?

 

[Mentallic]

Yes, because to go the other way, that is, take the derivative of the result $$\frac{ln|x^2+x|}{2x+1}$$ you need to use the quotient rule. It's not as simple as treating 2x+1 as a constant, which is what you instead get if the function in the log is linear.

e.g.

$$\int\frac{1}{ax}dx=\frac{ln|ax|}{a}$$

$$\frac{d}{dx}\left(\frac{ln|ax|}{a}\right)=\frac{1}{ax}\frac{a}{a}=\frac{1}{ax}$$

While

$$\int\frac{1}{ax^2}dx \neq \frac{ln|ax^2|}{2ax}$$

$$\frac{d}{dx}\left(\frac{ln|ax^2|}{2ax}\right)=\frac{\frac{1}{ax^2}.2ax-2a.ln|ax^2|}{4a^2x^2} \neq \frac{1}{ax^2}$$ as required.

 

[thereddevils]

Yes, because to go the other way, that is, take the derivative of the result $$\frac{ln|x^2+x|}{2x+1}$$ you need to use the quotient rule. It's not as simple as treating 2x+1 as a constant, which is what you instead get if the function in the log is linear.

e.g.

$$\int\frac{1}{ax}dx=\frac{ln|ax|}{a}$$

$$\frac{d}{dx}\left(\frac{ln|ax|}{a}\right)=\frac{1}{ax}\frac{a}{a}=\frac{1}{ax}$$

While

$$\int\frac{1}{ax^2}dx \neq \frac{ln|ax^2|}{2ax}$$

$$\frac{d}{dx}\left(\frac{ln|ax^2|}{2ax}\right)=\frac{\frac{1}{ax^2}.2ax-2a.ln|ax^2|}{4a^2x^2} \neq \frac{1}{ax^2}$$ as required.

thanks , so it only works when the denominator is a linear function .