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Natural Log integrals

2 methods of integrating the following:
Unparseable latex formula:

[br] & \int{\frac{1}{4}\left( \frac{1}{1+x} \right)}dx=\frac{1}{4}\ln (1+x)\,\,\,\,\,\,\,\,\,\,\,correct \\ [br] & \int{\left( \frac{1}{4+4x} \right)}dx=\frac{1}{4}\int{\left( \frac{4}{4+4x} \right)}\,dx=\frac{1}{4}ln(4+4x)\,\,\,\,wrong \\ [br]\end{align}



Why does the 2nd method not give the correct result?
Reply 1
Avatar for nerak99
nerak99
13
Well actually the second is only different from the first by a constant and so it is still correct.

14ln⁡4(1+x)=14[ln⁡4+ln⁡(1+x)]=14ln⁡4+14ln⁡(1+x)=14ln⁡4+\frac{1}{4}\ln 4(1+x)=\frac{1}{4}[\ln 4 +\ln (1+x)]=\frac{1}{4}\ln 4 + \frac{1}{4}\ln(1+x) = \frac{1}{4}\ln 4+ your first answer.

i.e they only differ by a constant. Indefinite integrals have a +c on the end.
(edited 8 years ago)
Original post by xlaser31
2 methods of integrating the following:
Unparseable latex formula:

[br] & \int{\frac{1}{4}\left( \frac{1}{1+x} \right)}dx=\frac{1}{4}\ln (1+x)\,\,\,\,\,\,\,\,\,\,\,correct \\ [br] & \int{\left( \frac{1}{4+4x} \right)}dx=\frac{1}{4}\int{\left( \frac{4}{4+4x} \right)}\,dx=\frac{1}{4}ln(4+4x)\,\,\,\,wrong \\ [br]\end{align}



Why does the 2nd method not give the correct result?


The constant will be different.
Original post by xlaser31
2 methods of integrating the following:
Unparseable latex formula:

[br] & \int{\frac{1}{4}\left( \frac{1}{1+x} \right)}dx=\frac{1}{4}\ln (1+x)\,\,\,\,\,\,\,\,\,\,\,correct \\ [br] & \int{\left( \frac{1}{4+4x} \right)}dx=\frac{1}{4}\int{\left( \frac{4}{4+4x} \right)}\,dx=\frac{1}{4}ln(4+4x)\,\,\,\,wrong \\ [br]\end{align}



Why does the 2nd method not give the correct result?


1. You are making trouble for yourself by omitting the constant of integration. It's important here.

2. Consider the rules of logarithms w.r.t to products, and point 1. above.

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