indefinite integral using partial fractions

Question

∫(x2-3x-7)/(2x+3)(x+1)²dx

Answer 1

Using partial fractions x² -3x -7/(2x+3)(x+1)² becomes A/(2x+3) + B /(x+1) +C/(x+1)². Hence x² -3x -7 = A(x+1)² + B(2x+3)(x+1) + C(2x+3). Matching coefficients gives 1= A+ 2B, -3= 2A +5B+2C, -7= A+ 3B +3C. Solving this set of equations using the substitution method gives A= 5/3 B= -1/3 and C= -23/6. So the integral now becomes 5/3∫dx/(2x+3) -1/3∫dx/(x+1) -23/6∫dx/(x+1)². The first term gives 5/6 ln|2x+3| second term is -1/3 ln|x+1| and the last term is +23/6(x+1) where the (x+1) is in the denominator.

Answer 2

desmos.com/calculator/dkvbtqrk4v