I am not sure if I am integrating this correctly. The constant in the numerator is throwing me off a bit. integrate 3/(2x+5)dx =3(1/x^2+5x+C)
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Originally Posted by startingover I am not sure if I am integrating this correctly. The constant in the numerator is throwing me off a bit. integrate 3/(2x+5)dx Hint: $\displaystyle \int \frac{3}{2x+5} = 3\int \frac{1}{2x+5} dx$
Originally Posted by startingover I am not sure if I am integrating this correctly. The constant in the numerator is throwing me off a bit. integrate 3/(2x+5)dx =3(1/x^2+5x+C) $\displaystyle \int \frac {3}{2x + 5}dx = 3 \int \frac{1}{2x + 5} dx$ Now use substitution, $\displaystyle u = 2x + 5$ Can you take it from here?
Originally Posted by startingover integrate 3/(2x+5)dx $\displaystyle \int\frac3{2x+5}~dx=3\int\frac1{2x+5}~dx=\frac32\i nt\frac{(2x+5)'}{2x+5}~dx=\frac32\ln\left|2x+5\rig ht|+k$
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