about two method that don't produce same result

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hyourinn

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It's about integral of x/(x+1) dx.
It's common that people change it form into
x+1-1/(x+1 then 1 - 1/(x+1)
and if integrated it became
x - ln |x+1| +C ←←←←←←result 1
I'm curious with this equation since when i'm using trigonometic subtitution for integrating this one. and both doesn't display same result. so i will start again
∫ x/(x+1) dx i subtitute x by tan²θ and dx by 2tanθsec²θ dθ
it will turn
∫(tan²θ/(tan²θ+1) 2tanθsec²θ dθ
∫(tan²θ/sec²θ) 2tanθsec²θ dθ
simplifying
∫2tan³θ dθthen
2∫tanθ tan²θ dθ
2∫tanθ (sec²θ-1) dθ
2∫tanθsec²θ-tanθ dθ
「for tanθsec²θ, i use subtitution tanθ=y sec²θdθ=dy so y dy = ½y²+ c subtitute back
y²=tan²θ=x」that conclude the first part,i rewrite it
2(½x - ∫ tanθ dθ) and for second part it's well known integral that ∫tan = -ln |cos| + c and since
tan²θ=x,
tanθ=√x so we know that value of cosθ is √(x+1)
finally
∫x/x+1 = x + ½ ln|√(x+1)| + C ←←←←←← result2

well it's very different so i would really glad if someone can explain to me, i'm fully aware that sometimes an equation need to be solved by special method instead of regular ones, but i don't think this one is included into that category.
thanks in advance
 
Last edited by a moderator:
hyourinn said:
It's about integral of x/x+1 dx.
It's common that people change it form into
x+1-1/x+1 then 1 - 1/x+1
and if integrated it became
x - ln |x+1| +C ←←←←←←result 1
I'm curious with this equation since when i'm using trigonometic subtitution for integrating this one. and both doesn't display same result. so i will start again
∫ x/x+1 dx i subtitute x by tan²θ and dx by 2tanθsec²θ dθ
it will turn
∫(tan²θ/tan²θ+1) 2tanθsec²θ dθ
∫(tan²θ/sec²θ) 2tanθsec²θ dθ
simplifying
∫2tan³θ dθthen
2∫tanθ tan²θ dθ
2∫tanθ (sec²θ-1) dθ
2∫tanθsec²θ-tanθ dθ
「for tanθsec²θ, i use subtitution tanθ=y sec²θdθ=dy so y dy = ½y²+ c subtitute back
y²=tan²θ=x」that conclude the first part,i rewrite it
2(½x - ∫ tanθ dθ) and for second part it's well known integral that ∫tan = -ln |cos| + c ....no neg in front of the ln here
and since tan²θ=x,tanθ=√x so we know that value of cosθ is
√(x+1)
finally
∫x/x+1 = x + ½ ln|√(x+1)| + C ←←←←←← result2 should be x - ...

well it's very different so i would really glad if someone can explain to me, i'm fully aware that sometimes an equation need to be solved by special method instead of regular ones, but i don't think this one is included into that category.
thanks in advance
Click to expand...

Recall the rule a log x = log (x^a).

If you fix up your sign and apply the log rule above, the results are the same.
 
hyourinn said:
It's about integral of x/x+1 dx.
It's common that people change it form into
x+1-1/x+1 then 1 - 1/x+1
and if integrated it became
x - ln |x+1| +C ←←←←←←result 1
I'm curious with this equation since when i'm using trigonometic subtitution for integrating this one. and both doesn't display same result. so i will start again
∫ x/x+1 dx i subtitute x by tan²θ and dx by 2tanθsec²θ dθ
it will turn
∫(tan²θ/tan²θ+1) 2tanθsec²θ dθ
∫(tan²θ/sec²θ) 2tanθsec²θ dθ
simplifying
∫2tan³θ dθthen
2∫tanθ tan²θ dθ
2∫tanθ (sec²θ-1) dθ
2∫tanθsec²θ-tanθ dθ
「for tanθsec²θ, i use subtitution tanθ=y sec²θdθ=dy so y dy = ½y²+ c subtitute back
y²=tan²θ=x」that conclude the first part,i rewrite it
2(½x - ∫ tanθ dθ) and for second part it's well known integral that ∫tan = -ln |cos| + c and since tan²θ=x,tanθ=√x so we know that value of cosθ is
1/√(x+1) <-- you meant this, right?
finally
∫x/x+1 = x + ½ ln|√(x+1)| + C ←←←←←← result2 <-- did you mean 2 rather than 1/2?

well it's very different so i would really glad if someone can explain to me, i'm fully aware that sometimes an equation need to be solved by special method instead of regular ones, but i don't think this one is included into that category.
thanks in advance
Click to expand...

I think you just made two errors in the second method. Fix those, and it should work out.
 
thanks everyone for help
and for subhotosh kahn, i little confuse why you said that no - infront of ln .
i'm pretty sure integral of tan = - ln |cos| and infront of tan were negative to begin with(after split tan² into sec²-1) so it became - -ln |cos| or i write as ln |cos| so could you help me understand bit more?
 
Last edited:
Dr.Peterson said:
I think you just made two errors in the second method. Fix those, and it should work out.
Click to expand...
surely i'm wrong in the second point. thanks for correcting it, additionally can you tell me how cosθ turned into 1/(√(x+1)) instead of √(x+1) ?
 
hyourinn said:
thanks everyone for help
and for subhotosh kahn, i little confuse why you said that no - infront of ln .
i'm pretty sure integral of tan = - ln |cos| and infront of tan were negative to begin with(after split tan² into sec²-1) so it became - -ln |cos| or i write as ln |cos| so could you help me understand bit more?
Click to expand...
Sorry, my mistake, you are correct at that step! Disregard everything I said.

(Actually it's Harry_the_cat, but I'll let Subhotosh take the blame.)
 
hyourinn said:
surely i'm wrong in the second point. thanks for correcting it, additionally can you tell me how cosθ turned into 1/(√(x+1)) instead of √(x+1) ?
Click to expand...
One way:
tan(x) = x^(1/2)

tan2(x) = x

1 + tan2(x) = x + 1

sec2(x) = x + 1

1/cos2(x) = x + 1

cos2(x) = 1/(x + 1)...... and continue.....
 
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