Thread: Need help with integration (questions with fractions, and e)

1. Need help with integration (questions with fractions, and e)

Hi, I can do basic integration but on stuck on a few questions, with fractions and e. If you show me how to do these two, I'm sure I could do the rest.

1. dx

2. dx

Thanks, really appreciate any input.

2. Re: Need help with integration (questions with fractions, and e)

Hello

1. $\displaystyle \mathrm{\displaystyle\int\dfrac{1}{4x+1}dx=\dfrac{ \ln(4x+1)}{4}+C}$

$\displaystyle \boxed{\mathrm{\displaystyle\int\dfrac{dt}{t}=\ln t+C}}$

2. $\displaystyle \mathrm{\displaystyle\int4e^{1-4x}=-e^{1-4x}+C}$

$\displaystyle \boxed{\mathrm{\displaystyle\int e^t=e^t+C}}$

Greetings

3. Re: Need help with integration (questions with fractions, and e)

Originally Posted by darthjavier
Hello

1. $\displaystyle \mathrm{\displaystyle\int\dfrac{1}{4x+1}dx=\dfrac{ \ln(4x+1)}{4}+C}$

$\displaystyle \boxed{\mathrm{\displaystyle\int\dfrac{dt}{t}=\ln t+C}}$

2. $\displaystyle \mathrm{\displaystyle\int4e^{1-4x}=-e^{1-4x}+C}$

$\displaystyle \boxed{\mathrm{\displaystyle\int e^t=e^t+C}}$

Greetings
No, the first is incorrect. It should be \displaystyle \displaystyle \begin{align*} \int{\frac{1}{4x + 1}\,dx} = \frac{\ln{|4x + 1|}}{4} + C \end{align*} since \displaystyle \displaystyle \begin{align*} \int{\frac{1}{t}\,dt} = \ln{|t|} + C \end{align*}.

4. Re: Need help with integration (questions with fractions, and e)

Originally Posted by Prove It
No, the first is incorrect. It should be \displaystyle \displaystyle \begin{align*} \int{\frac{1}{4x + 1}\,dx} = \frac{\ln{|4x + 1|}}{4} + C \end{align*} since \displaystyle \displaystyle \begin{align*} \int{\frac{1}{t}\,dt} = \ln{|t|} + C \end{align*}.
Oops I'm sorry :P