Final Answer
$\frac{y}{2y-1}+\frac{-1}{2\left(2y-1\right)}+\frac{1}{2}\ln\left(2y-1\right)+C_0$
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Step-by-step Solution
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1
Rewrite the fraction $\frac{1}{2y-1}$ inside the integral as the product of two functions: $1\left(\frac{1}{2y-1}\right)$
$\int1\left(\frac{1}{2y-1}\right)dy$
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$\int1\left(\frac{1}{2y-1}\right)dy$
Learn how to solve problems step by step online. Find the integral int(1/(2y-1))dy. Rewrite the fraction \frac{1}{2y-1} inside the integral as the product of two functions: 1\left(\frac{1}{2y-1}\right). We can solve the integral \int1\left(\frac{1}{2y-1}\right)dy by applying integration by parts method to calculate the integral of the product of two functions, using the following formula. First, identify u and calculate du. Now, identify dv and calculate v.
Final Answer
$\frac{y}{2y-1}+\frac{-1}{2\left(2y-1\right)}+\frac{1}{2}\ln\left(2y-1\right)+C_0$