Answer
$$\frac{5}{2}\ln \left| {2x - 1} \right| + 3\ln \left| {x + 4} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{11x + 17}}{{2{x^2} + 7x - 4}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& {\text{Factor the denominator}} \cr
& \frac{{11x + 17}}{{2{x^2} + 7x - 4}} = \frac{{11x + 17}}{{\left( {2x - 1} \right)\left( {x + 4} \right)}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{{11x + 17}}{{\left( {2x - 1} \right)\left( {x + 4} \right)}} = \frac{A}{{2x - 1}} + \frac{B}{{x + 4}} \cr
& {\text{Multiplying the equation by }}\left( {2x - 1} \right)\left( {x + 4} \right){\text{, we have}} \cr
& 11x + 17 = A\left( {x + 4} \right) + B\left( {2x - 1} \right) \cr
& {\text{if we set }}x = 1/2, \cr
& 11\left( {\frac{1}{2}} \right) + 17 = A\left( {\frac{1}{2} + 4} \right) + B\left( 0 \right) \cr
& A = 5 \cr
& {\text{if we set }}x = - 4, \cr
& 11\left( { - 4} \right) + 17 = A\left( 0 \right) + B\left( {2\left( { - 4} \right) - 1} \right) \cr
& B = 3 \cr
& {\text{then}} \cr
& \frac{{11x + 17}}{{\left( {2x - 1} \right)\left( {x + 4} \right)}} = \frac{5}{{2x - 1}} + \frac{3}{{x + 4}} \cr
& \cr
& \int {\frac{{11x + 17}}{{2{x^2} + 7x - 4}}} dx = \int {\left( {\frac{5}{{2x - 1}} + \frac{3}{{x + 4}}} \right)dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{2}\int {\frac{2}{{2x - 1}}dx} + 3\int {\frac{1}{{x + 4}}} dx \cr
& {\text{integrating}} \cr
& = \frac{5}{2}\ln \left| {2x - 1} \right| + 3\ln \left| {x + 4} \right| + C \cr} $$
