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

**EarlyTravel** · November 13th 2012, 02:29 AM

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.

**darthjavier** · November 13th 2012, 03:43 AM

Hello

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

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

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

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

Greetings

**Prove It** · November 13th 2012, 03:56 AM

Originally Posted by darthjavier

Hello

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

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

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

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

Greetings

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

**darthjavier** · November 13th 2012, 06:12 AM

Originally Posted by Prove It

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

Oops I'm sorry :P

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