LN Integrals

[palangi]

I understand that the anti-derivative of 1/x is lnx. What are the rules though? If I am evaluating intergal 1/x+2, or 1/sinx, or 1/x^2 are the answers ln|x+2|, ln|sinx|, ln|x^2|

What I think I understand the pattern to be is that I can use ln as long as there is not any exponents in the denominator (that would make my example of integral 1/x^2 incorrect), is this accurate?

 

[galactus]

I understand that the anti-derivative of 1/x is lnx. What are the rules though? If I am evaluating intergal 1/x+2, or 1/sinx, or 1/x^2 are the answers ln|x+2|, ln|sinx|, ln|x^2|

I am sorry, it does not always work that way.

For \(\displaystyle \int\frac{1}{x+2}=ln(x+2), \;\ \;\ \int\frac{1}{sin(x)}dx=\int csc(x)dx=ln|csc(x)-cot(x)|=ln|tan(\frac{x}{2})|, \;\ \;\ \int\frac{1}{x^{2}}dx=\int x^{-2}dx=\frac{-1}{x}\)

But, if you have something like this: \(\displaystyle \int\frac{1}{ax+b}dx=\frac{ln(ax+b)}{a}\)

 

[palangi]

couldn't just make it easy for me huh?

Okay then, what I have is Integral (I have no idea how to make the integral symbol you do) of x/(1+x^2)^(1/2)

I could easily solve this without the square root, but... I think I can use a u substitution, but would I use it for just 1+x^2 and then have 1/2lnx^(1/2). It seems wrong to me though. The other option I was thinking is make the square root a negative exponent and solve that way, but that seems wrong too.

Your help really is appreciated.

 

[galactus]

Let \(\displaystyle u=\sqrt{1+x^{2}}, \;\ du=\frac{x}{\sqrt{x^{2}+1}}dx\)

Now, it's an easy integration.

All your left with is \(\displaystyle \int du\)

 

[palangi]

Even I can do that one. You're wonderful. Thank you!