Problem with a Definite Integral

Question

I'm studying for finals and having issues with a question about the Fundamental Theorem of Calculus. The question is: $$\int_1^6 \frac {\mathrm{d}t}{4t+23}.$$
I took the integral of $$\frac{\mathrm{d}t}{4t+23}$$ and got $$\ln\left|4t+23\right|$$ I then used Fundamental Theorem of Calculus and plugged in $$F(6)-F(1)$$ and got $$\ln(47)-\ln(27)$$ But the program I'm using has it marked wrong. Can you help me find my mistake ?

Answer 1

You are wrong by a factor: $$\frac{d}{dt}\ln(4t+23) = \frac{4}{4t+23},$$ so the correct antiderivative is $$\frac{\ln(4t+23)}{4}.$$

Answer 2

HINT: Remember in general $$\int \frac{dx}{ax+b}=\frac{1}{a}\ln|ax+b|+C$$ hence, one should get $$\int_{1}^{6} \frac{dt}{4t+23}=\frac{1}{4}[\ln|4t+23|]_{1}^6$$

Answer 3

$$\int_{1}^{6}\frac{1}{4t+23}\space\text{d}t=$$

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Substitute $u=4t+23$ and $\text{d}u=4\space\text{d}t$.

Then the new lower bound is $u=23+4\cdot1=27$ and upper bound $u=23+4\cdot6=47$:

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$$\frac{1}{4}\int_{27}^{47}\frac{1}{u}\space\text{d}u=$$ $$\frac{1}{4}\left[\ln\left|u\right|\right]_{27}^{47}=$$ $$\frac{1}{4}\left(\ln\left|47\right|-\ln\left|27\right|\right)=$$ $$\frac{1}{4}\left(\ln\left(\frac{47}{27}\right)\right)=\frac{\ln\left(\frac{47}{27}\right)}{4}$$