Final answer to the problem
Step-by-step Solution
Specify the solving method
Find the integral
Expand the integral $\int\left(\frac{x}{4x^2+2x}+\frac{-3}{2x+1}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately
We can solve the integral $\int\frac{-3}{2x+1}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2x+1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above
Isolate $dx$ in the previous equation
Substituting $u$ and $dx$ in the integral and simplify
The integral $\frac{1}{2}\int\frac{-3}{u}du$ results in: $-\frac{3}{2}\ln\left(2x+1\right)$
Gather the results of all integrals
Rewrite the expression $\frac{x}{4x^2+2x}$ inside the integral in factored form
We can solve the integral $\int\frac{1}{2\left(2x+1\right)}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2x+1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above
Isolate $dx$ in the previous equation
Substituting $u$ and $dx$ in the integral and simplify
The integral $\frac{1}{4}\int\frac{1}{u}du$ results in: $\frac{1}{4}\ln\left(2x+1\right)$
Gather the results of all integrals
Combining like terms $\frac{1}{4}\ln\left(2x+1\right)$ and $-\frac{3}{2}\ln\left(2x+1\right)$
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$