Taking factors out of this integral?

Question

In the integral: $$\int \left( \frac{-25}{17(2t+3)} + \frac{37}{17(5t-1)} \right) dt$$ Why is the final answer: $$-\frac{25}{34} \ln|2t + 3| + \frac{37}{85} \ln|5t - 1| + C$$

If you take $-\frac{1}{2}$ as well as the $-\frac{25}{17}$ doesn't the fraction become: $$\frac{1}{t + 3/2}$$

And if you as well take the $\frac{1}{5}$ out of the other fraction shouldn't it become: $$\frac{1}{t - 1/5}$$

I would have got a final answer of: $$-\frac{25}{34} \ln \left| t + \frac{3}{2} \right| + \frac{37}{85} \ln \left| t - \frac{1}{5} \right| + C$$

Answer 1

This is a common misunderstanding around integrals resulting in logarithms. Both answers are correct and equivalent.

$$-\frac{25}{34}\ln|2t + 3| + \frac{37}{85}\ln|5t - 1| + C_1$$

$$-\frac{25}{34}\ln|t + 3/2| + \frac{37}{85}\ln|t - 1/5| + C_2$$

To understand why note that: $$\ln|5t-1|=\ln|5(t-1/5)|=\ln5+\ln|t-1/5|$$ and $$\ln|2t+3|=\ln|2(t+3/2)|=\ln2+\ln|t+3/2|$$

So the two answers only differ by the constant $\frac{25}{34}\ln2+\frac{37}{85}\ln5$.

In other words the two constants of integration are not numerically equal and differ by this amount.

Answer 2

$\int\frac{-25}{17(2t+3)} + \frac{37}{17(5t-1)} dt$ proceeds to be

$-\frac{25}{17}\int\frac{1}{(2t+3)}dt + \frac{37}{17}\int\frac{1}{(5t-1)} dt$

for the first integral, take $u = 2t+ 3$ so $du = 2 dt$. Then you get

$-\frac{25}{34}\int\frac{du}{(u)}$

for the second integral, take $u = 5t - 1$ so $du = 5 dt$. Then you get

$\frac{37}{85}\int\frac{du}{(u)}$

HINT

If you were to integrate the first integral, you would get $ln(u) + C = ln(2(t+\frac{3}{2})) = ln(2) + ln(t+\frac{3}{2}) + C$

Answer 3

You cannot just "factor out" 2 from $ln(2t+3)$. You only can factor out exponents: $ln(a^x)=x\cdot ln(a)$

The integral of $\frac{1}{at+b}$ is $\frac{1}{a}\cdot ln(at+b)+C$

The function $g(t)$ of $ln(g(t))$ is always the denominator of the derivative. But if you use the chain rule $g(t)$ has also be differentiated. This $g'(t)$ has to be neutralized.

The derivative of $\frac{1}{2}\cdot ln(\underbrace{2t+3}_{g(t)})+C$ is equal to

$\frac{1}{2}\cdot\frac{1}{2t+3}\cdot \underbrace{2}_{g'(t)}$

Answer 4

$$\int\left(\frac{-25}{17(2t+3)}+\frac{37}{17(5t-1)}\right)\space\text{d}t=$$ $$\int\frac{-25}{17(2t+3)}\space\text{d}t+\int\frac{37}{17(5t-1)}\space\text{d}t=$$ $$-\frac{25}{17}\int\frac{1}{2t+3}\space\text{d}t+\frac{37}{17}\int\frac{1}{5t-1}\space\text{d}t=$$

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• Substitute $u=2t+3$ and $\text{d}u=2\space\text{d}t$.
• Substitute $s=5t-1$ and $\text{d}s=5\space\text{d}t$.

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$$-\frac{25}{34}\int\frac{1}{u}\space\text{d}u+\frac{37}{85}\int\frac{1}{s}\space\text{d}s=$$ $$-\frac{25\ln\left|u\right|}{34}+\frac{37\ln\left|s\right|}{85}+\text{C}=$$ $$-\frac{25\ln\left|2t+3\right|}{34}+\frac{37\ln\left|5t-1\right|}{85}+\text{C}=$$ $$\frac{37\ln\left|5t-1\right|}{85}-\frac{25\ln\left|2t+3\right|}{34}+\text{C}$$