Antiderivative

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SB28210

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Aug 6, 2009
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Good evening:

I'm trying to solve find the antiderivative for this function evaluated from zero to 5:

.08 + .025X/(X + 1) dx.

the answer is = 0.4802

...but the X/(X+1) part is giving me fits. I can't think of a substitution for this (neither the numerator not the denominator is a derivative or the other) and when I try multiplying by X + 1 to clear the denominator to get

.08X + .08 + .025X dx does not give me the right answer when I integrate.

I would appreciate someone steering me in the right direction on this one.

Kind regards,

SteveB
 
\(\displaystyle \int_{0}^{5}\bigg[.08+\frac{.025x}{x+1}\bigg]dx \ = \ \int_{0}^{5}.08dx+\int_{0}^{5}\frac{.025x}{x+1}dx\)

\(\displaystyle = \ .08x\bigg]_{0}^{5}, \ let \ u \ = \ x+1, \ x \ = \ u-1 \ and \ du \ = \ dx\)

\(\displaystyle = \ .4 \ +\int_{1}^{6}\frac{.025u-.025}{u}du \ = \ .4+\int_{1}^{6}[.025-.025u^{-1}]du\)

\(\displaystyle =.4+\bigg[.025u-.025ln|u|\bigg]_{1}^{6} \ = \ .4+.15-.025ln(6)-.025 \ \dot= \ .4802\)
 
SB28210 said:
Good evening:

I'm trying to solve find the antiderivative for this function evaluated from zero to 5:

.08 + .025X/(X + 1) dx.

the answer is = 0.4802

...but the X/(X+1) part is giving me fits. I can't think of a substitution for this (neither the numerator not the denominator is a derivative or the other) and when I try multiplying by X + 1 to clear the denominator to get

.08X + .08 + .025X dx does not give me the right answer when I integrate.

I would appreciate someone steering me in the right direction on this one.

Kind regards,

SteveB
Click to expand...

\(\displaystyle \frac{x}{x+1} \ = \ \ 1 \ - \ \frac{1}{x+1}\)

Now antiderivative should be simple.....
 
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