Question

$\int \frac{2x+3}{5x-1} dx$

Answer 1

$\frac{2}{5}x + \frac{17}{25} \ln|5x-1| + C$

Answer 2

To integrate the given function, we will use the method of adjusting the numerator to be a multiple of the denominator plus a constant.

Let the given integral be $I = \int \frac{2x+3}{5x-1} dx$.

First, we express the numerator ($2x+3$) in terms of the denominator ($5x-1$). We want to find constants A and B such that: $2x+3 = A(5x-1) + B$

Expand the right side: $2x+3 = 5Ax - A + B$

Now, compare the coefficients of $x$ and the constant terms on both sides: Comparing coefficients of $x$: $5A = 2 \implies A = \frac{2}{5}$

Comparing constant terms: $-A + B = 3$ Substitute the value of $A$: $-\frac{2}{5} + B = 3$ $B = 3 + \frac{2}{5}$ $B = \frac{15+2}{5} = \frac{17}{5}$

Now substitute these values of A and B back into the numerator expression: $2x+3 = \frac{2}{5}(5x-1) + \frac{17}{5}$

Substitute this back into the integral: $I = \int \frac{\frac{2}{5}(5x-1) + \frac{17}{5}}{5x-1} dx$

Separate the terms in the numerator: $I = \int \left( \frac{\frac{2}{5}(5x-1)}{5x-1} + \frac{\frac{17}{5}}{5x-1} \right) dx$ $I = \int \left( \frac{2}{5} + \frac{17}{5(5x-1)} \right) dx$

Now, integrate term by term: $I = \int \frac{2}{5} dx + \int \frac{17}{5(5x-1)} dx$ $I = \frac{2}{5} \int 1 dx + \frac{17}{5} \int \frac{1}{5x-1} dx$

The first integral is straightforward: $\int 1 dx = x$

For the second integral, $\int \frac{1}{5x-1} dx$, we use a substitution. Let $u = 5x-1$. Then, differentiate $u$ with respect to $x$: $\frac{du}{dx} = 5$ $du = 5 dx \implies dx = \frac{1}{5} du$

Substitute $u$ and $dx$ into the integral: $\int \frac{1}{u} \left(\frac{1}{5} du\right) = \frac{1}{5} \int \frac{1}{u} du$ We know that $\int \frac{1}{u} du = \ln|u| + C'$ So, $\frac{1}{5} \ln|u| + C' = \frac{1}{5} \ln|5x-1| + C'$

Now, combine all parts of the integral: $I = \frac{2}{5}x + \frac{17}{5} \left( \frac{1}{5} \ln|5x-1| \right) + C$ $I = \frac{2}{5}x + \frac{17}{25} \ln|5x-1| + C$ where C is the constant of integration.