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I'm trying to solve the integral below:

$$\int_2^8 \frac{dt}{4t+14}$$

The issue for me is not calculating the integral, but the antiderivative. My steps for calculating it looked something like this:

  1. $\displaystyle\int_2^8 \frac{1}{2} \frac{dt}{2t+7}$
  2. $\displaystyle\frac{1}{2}\int_2^8 \frac{dt}{2t+7}$
  3. $\displaystyle\frac{1}{2}\ln(2t + 7)$

I double checked my answer with Wolfram|Alpha, but it (correctly) calculated it as:

$$\frac{1}{4}\ln(2t + 7)$$

I looked at the steps it took, but they don't make much sense for me. Hopefully someone can help me out with this.

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When you found the antiderivative of $ln(2t+7)$ you forgot to use the reversed chain rule to obtain another factor of $1/2$. –  Ravi Donepudi Jan 26 '12 at 14:23
    
Once again: "solve" is the wrong word here. You're trying to evaluate the integral. –  Michael Hardy Jan 26 '12 at 15:53

3 Answers 3

up vote 4 down vote accepted

Your first two step looks good, but you made a mistake at 3rd. You could continue like the following:

$$\frac{1}{2}\cdot\frac{1}{2} \int \frac{2dt}{2t+7}=\frac{1}{4}\ln(2t+7)$$

You probably forgot to take into account the derivative of $2t+7$.

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Check your answer: you claim $\ln(2t+7)$ is an antiderivative of ${1\over 2t+7}$. Is this correct? Check: $$ {d\over dt} \ln(2t+7)= {1\over 2t+7}\cdot 2 \ne{{1\over 2t+7} } $$ So, no...

As mentioned in the comments, you forgot about the chain rule... $\int {1\over 2t+7}\,dt$ "looks like" $\int {1\over u}\,du$; so, you might use a substitution:

$$\int {1\over 2t+7}\, dt\ \ \buildrel {u=2t+7}\over {=}\ \ \int {1\over u}\cdot {1\over2}\,du={1\over2} {\ln|u|}+C={1\over2}\ln|2x+7|+C. $$

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When you're going from $\int\frac{dt}{2t+7}$ to $\log(2t+7)$, you're probably thinking about the fact that $\int\frac{dx}{x}=\log|x|+c$. Let's let $u=2t+7$, which means that $du=2dt$ or $dt=\frac{1}{2}du$. (This technique is sometimes called substitution or $u$-substitution.) Now, $$\int\frac{dt}{2t+7}=\int\frac{\frac{1}{2}du}{u}=\frac{1}{2}\log|u|+c=\cdots$$

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