Решите сложные интегралы (Фото ниже)

1. ∫(3x+4)/(x²+3x+4)dx=∫(((3(2x+3))/(2(x²+3x+4)))-(17)/(2(x²+3x+4))))dx=3/2∫(2x+3)/(x²+3x+4)dx-17/2∫1/(x²+3x+4)dx= 3/2∫1/udu-17/2∫1/((x+3/2)²+7/4)dx=(3lnu)/2-17/2∫1/(s²+7/4)ds=(3lnu)/2-17/2∫4/(7((4s²/7)+1))ds=(3lnu)/2-34/7∫1/((4s²/7)+1)ds=(3lnu)/2-17/√7∫1/(p²+1)dp=(3lnu)/2-(17arctgp)/√7+C=(3lnu)/2-(17arctg(2s/√7))/√7+C=(3ln(x²+3x+4))/2-(17arctg((2x+3))/√7))/√7+C u=x²+3x+4 du=(2x+3)dx s=x+3/2 ds=dx p=2s/√7 dp=2/√7ds 2. ∫(4x-3)/(x³+4x²-x-4)dx=∫((7/(6(x+1)))-(19/(15(x+4)))+(1/(10(x-1))))dx=7/6∫1/(x+1)dx-19/15∫1/(x+4)dx+1/10∫1/(x-1)dx=7/6∫1/udu-19/15∫1/sds+1/10∫1/pdp=7lnu/6-19lns/15+lnp/10+C=7/6*ln(x+1)-19*15*ln(x+4)+ln(x-1)/10+C u=x+1 du=dx s=x+4 ds=dx p=x-1 dp=dx