# How to evaluate the definite integral?

How to evaluate the definite integral?

$$\int \frac{7}{3x+1}dx$$ I am having difficulties to finish the question: Below is what I did: $$=\left.\frac{7}{3}\ln|3x+1|\right|_0^4$$ $$=\frac{7}{3}\ln(\dots.$$

• Please learn to use mathjax to format your mathematics more legibly. – rurouniwallace Jul 15 '13 at 20:09
• You're almost there...! – DonAntonio Jul 15 '13 at 20:09

All you have left to do is evaluate $$\dfrac 73 \ln|3x + 1| \Big|_0^4 = \dfrac 73 (\ln(13) - \underbrace{\ln(1)}_{\large = 0}) = \quad \frac 73 \ln(13) \quad \approx \quad5.985$$
• Yes, we evaluate the integral at $x = 13$ and subtract its evaluation at $x = 0$. $x = 4 \implies \frac 73\ln(3x + 1) = \frac 73 \ln(3\cdot 4 + 1) = \frac 73 \ln(13)$. At x = 0, we have $\frac 73\ln(3\cdot 0 + 1) = \frac 73\ln(1) = 0$ – amWhy Jul 15 '13 at 20:23
Hint:take $u=3x+1\to du=3dx$$\int_0^4\frac{7}{3x+1}dx=\frac73\int_0^4\frac{3dx}{3x+1}=\frac73\int_0^4\frac{du}{u}=lnu|_1^{13}=\frac73(\ln(13)-ln1)=\frac73(\ln(13)=\frac13\log_e {13^7}$$ Let's see if I guessed correctly what you wrote (use LaTeX for mathematics in this site, please!): $$\int\limits_0^4\frac7{3x+1}dx=\frac73\int\limits_0^4\frac{3\,dx}{3x+1}=\left.\frac73\log(3x+1)\right|_0^4=\frac73\left(\log13-\log1\right)=\ldots$$ • I came with y= 1.536 which is wrong. Can you tell me your answer? – Sara Sharp Jul 15 '13 at 20:18 • You have it there: $$\frac73\log 13\cong 5.9849$$ but I think it looks nicer as the left side... Also, why do you put that$\;y\;\$ there?? – DonAntonio Jul 15 '13 at 20:20