This integral has me stumped

[IAteYourMother]

finding the integral of

(Ln[1+x])/x^2 by hand.

 

[chuckywang]

Integrate by parts

 

[imported_Tick]

Originally posted by: chuckywang
Integrate by parts

And you need to do a U substitution on the LN.

 

[chuckywang]

Let du = 1/x^2, so u=-1/x

Let v = ln(1+x), so dv=1/(1+x)

Therefore, the integral becomes -ln(1+x)/x - Integral(1/(x*(1+x)dx)

You can expand that integral by partial fractions as: 1/x - 1/(1+x), so the integral is ln(x)-ln(1+x)

Therefore the original integral becomes: ln(1+x)(1-1/x)-ln(x)+C.

(Caveat: I did this without any checking, so check my work.)

 

[imported_Tick]

Originally posted by: chuckywang
Let du = 1/x^2, so u=-1/x

Let v = ln(1+x), so dv=1/(1+x)

Therefore, the integral becomes -ln(1+x)/x - Integral(1/(x*(1+x)dx)

You can expand that integral by partial fractions as: 1/x - 1/(1+x), so the integral is ln(x)-ln(1+x)

Therefore the original integral becomes: ln(1+x)(1-1/x)-ln(x)+C.

(Caveat: I did this without any checking, so check my work.)

What did you do here?

Why not: -ln(1+X)/x - Intergral(ln(x)) - Integral(ln(1+X)) ??

 

[chuckywang]

Originally posted by: Tick

Originally posted by: chuckywang
Let du = 1/x^2, so u=-1/x

Let v = ln(1+x), so dv=1/(1+x)

Therefore, the integral becomes -ln(1+x)/x - Integral(1/(x*(1+x)dx)

You can expand that integral by partial fractions as: 1/x - 1/(1+x), so the integral is ln(x)-ln(1+x)

Therefore the original integral becomes: ln(1+x)(1-1/x)-ln(x)+C.

(Caveat: I did this without any checking, so check my work.)

What did you do here?

Why not: -ln(1+X)/x - Intergral(ln(x)) - Integral(ln(1+X)) ??

Integral(1/(x*(1+x))dx) = Integral((1/x-1/(1+x))dx) = ln(x)-ln(1+x)-C

Therefore -ln(1+x)/x - Integral(1/(x*(1+x))dx) = -ln(1+x)/x - ln(x)+ln(1+x)+C = ln(1+x)(1-1/x)-ln(x)+C

 

[imported_Tick]

Originally posted by: chuckywang

Originally posted by: Tick

Originally posted by: chuckywang
Let du = 1/x^2, so u=-1/x

Let v = ln(1+x), so dv=1/(1+x)

Therefore, the integral becomes -ln(1+x)/x - Integral(1/(x*(1+x)dx)

You can expand that integral by partial fractions as: 1/x - 1/(1+x), so the integral is ln(x)-ln(1+x)

Therefore the original integral becomes: ln(1+x)(1-1/x)-ln(x)+C.

(Caveat: I did this without any checking, so check my work.)

What did you do here?

Why not: -ln(1+X)/x - Intergral(ln(x)) - Integral(ln(1+X)) ??

Integral(1/(x*(1+x))dx) = Integral((1/x-1/(1+x))dx) = ln(x)-ln(1+x)-C

Therefore -ln(1+x)/x - Integral(1/(x*(1+x))dx) = -ln(1+x)/x - ln(x)+ln(1+x)+C = ln(1+x)(1-1/x)-ln(x)+C

Oh, ok, not how I was gonna do it, but it looks good. I think... It's hard to read in this format.

 

[chuckywang]

Originally posted by: Tick

Originally posted by: chuckywang

Originally posted by: Tick

Originally posted by: chuckywang
Let du = 1/x^2, so u=-1/x

Let v = ln(1+x), so dv=1/(1+x)

Therefore, the integral becomes -ln(1+x)/x - Integral(1/(x*(1+x)dx)

You can expand that integral by partial fractions as: 1/x - 1/(1+x), so the integral is ln(x)-ln(1+x)

Therefore the original integral becomes: ln(1+x)(1-1/x)-ln(x)+C.

(Caveat: I did this without any checking, so check my work.)

What did you do here?

Why not: -ln(1+X)/x - Intergral(ln(x)) - Integral(ln(1+X)) ??

Integral(1/(x*(1+x))dx) = Integral((1/x-1/(1+x))dx) = ln(x)-ln(1+x)-C

Therefore -ln(1+x)/x - Integral(1/(x*(1+x))dx) = -ln(1+x)/x - ln(x)+ln(1+x)+C = ln(1+x)(1-1/x)-ln(x)+C

Oh, ok, not how I was gonna do it, but it looks good. I think... It's hard to read in this format.

Yeah, write it out on paper.

 

[hypn0tik]

Originally posted by: Tick

Originally posted by: chuckywang

Originally posted by: Tick

Originally posted by: chuckywang
Let du = 1/x^2, so u=-1/x

Let v = ln(1+x), so dv=1/(1+x)

Therefore, the integral becomes -ln(1+x)/x - Integral(1/(x*(1+x)dx)

You can expand that integral by partial fractions as: 1/x - 1/(1+x), so the integral is ln(x)-ln(1+x)

Therefore the original integral becomes: ln(1+x)(1-1/x)-ln(x)+C.

(Caveat: I did this without any checking, so check my work.)

What did you do here?

Why not: -ln(1+X)/x - Intergral(ln(x)) - Integral(ln(1+X)) ??

Integral(1/(x*(1+x))dx) = Integral((1/x-1/(1+x))dx) = ln(x)-ln(1+x)-C

Therefore -ln(1+x)/x - Integral(1/(x*(1+x))dx) = -ln(1+x)/x - ln(x)+ln(1+x)+C = ln(1+x)(1-1/x)-ln(x)+C

Oh, ok, not how I was gonna do it, but it looks good. I think... It's hard to read in this format.

How would you have done it?

 

[imported_Tick]

Originally posted by: chuckywang

Originally posted by: Tick

Originally posted by: chuckywang

Originally posted by: Tick

Originally posted by: chuckywang
Let du = 1/x^2, so u=-1/x

Let v = ln(1+x), so dv=1/(1+x)

Therefore, the integral becomes -ln(1+x)/x - Integral(1/(x*(1+x)dx)

You can expand that integral by partial fractions as: 1/x - 1/(1+x), so the integral is ln(x)-ln(1+x)

Therefore the original integral becomes: ln(1+x)(1-1/x)-ln(x)+C.

(Caveat: I did this without any checking, so check my work.)

What did you do here?

Why not: -ln(1+X)/x - Intergral(ln(x)) - Integral(ln(1+X)) ??

Integral(1/(x*(1+x))dx) = Integral((1/x-1/(1+x))dx) = ln(x)-ln(1+x)-C

Therefore -ln(1+x)/x - Integral(1/(x*(1+x))dx) = -ln(1+x)/x - ln(x)+ln(1+x)+C = ln(1+x)(1-1/x)-ln(x)+C

Oh, ok, not how I was gonna do it, but it looks good. I think... It's hard to read in this format.

Yeah, write it out on paper.

I'm not gonna hunt down a pen just to help with this guys homework.

 

[imported_Tick]

Originally posted by: hypn0tik

Originally posted by: Tick

Originally posted by: chuckywang

Originally posted by: Tick

Originally posted by: chuckywang
Let du = 1/x^2, so u=-1/x

Let v = ln(1+x), so dv=1/(1+x)

Therefore, the integral becomes -ln(1+x)/x - Integral(1/(x*(1+x)dx)

You can expand that integral by partial fractions as: 1/x - 1/(1+x), so the integral is ln(x)-ln(1+x)

Therefore the original integral becomes: ln(1+x)(1-1/x)-ln(x)+C.

(Caveat: I did this without any checking, so check my work.)

What did you do here?

Why not: -ln(1+X)/x - Intergral(ln(x)) - Integral(ln(1+X)) ??

Integral(1/(x*(1+x))dx) = Integral((1/x-1/(1+x))dx) = ln(x)-ln(1+x)-C

Therefore -ln(1+x)/x - Integral(1/(x*(1+x))dx) = -ln(1+x)/x - ln(x)+ln(1+x)+C = ln(1+x)(1-1/x)-ln(x)+C

Oh, ok, not how I was gonna do it, but it looks good. I think... It's hard to read in this format.

How would you have done it?

With magic. Seriously, I'm sick of trying to do by typing this out.