1. ∫ x 3 + 4 x + 3 x 4 + 5 x 2 + 4 d x = ∫ x 3 + 4 x + 3 ( x 2 + 1 ) ( x 2 + 4 ) d x = ∫ A x + B x 2 + 1 + C x + D x 2 + 4 d x x 3 + 4 x + 3 ( x 2 + 1 ) ( x 2 + 4 ) = A x + B x 2 + 1 + C x + D x 2 + 4 x 3 + 4 x + 3 ( x 2 + 4 ) = A x + B + C x + D x 2 + 4 ( x 2 + 1 ) x = i − i + 4 i + 3 3 = i + 1 = A i + B A = 1 B = 1 x 4 + 4 x 2 + 3 x x 4 + 5 x 2 + 4 = x 2 + x x 2 + 1 + C x 2 + D x x 2 + 4 x → ∞ 1 = 1 + C C = 0 x = 0 3 4 = 1 + D 4 D = − 1 ∫ x 3 + 4 x + 3 ( x 2 + 1 ) ( x 2 + 4 ) d x = ∫ x + 1 x 2 + 1 + − 1 x 2 + 4 d x = ∫ x x 2 + 1 d x + ∫ 1 x 2 + 1 d x + ∫ − 1 x 2 + 4 d x = ln ( x 2 + 1 ) 2 + tan − 1 x − 1 2 tan − 1 x 2 2. ∫ 1 x + x x d x = 2 ∫ 1 x + x d x = 2 ∫ ( A x + B x + 1 ) d x A = 1 B = − 1 = 2 ∫ ( 1 x + − 1 x + 1 ) d x = 2 ( ∫ 1 x d x − ∫ 1 x + 1 d x ) = 2 ( ln x − ln ( x + 1 ) ) = 2 ln x x + 1 {\displaystyle {\begin{aligned}&1.\;\int {\frac {x^{3}+4x+3}{x^{4}+5x^{2}+4}}dx\\&=\int {\frac {x^{3}+4x+3}{\left(x^{2}+1\right)\left(x^{2}+4\right)}}dx\\&=\int {\frac {Ax+B}{x^{2}+1}}+{\frac {Cx+D}{x^{2}+4}}dx\\&{\frac {x^{3}+4x+3}{\left(x^{2}+1\right)\left(x^{2}+4\right)}}={\frac {Ax+B}{x^{2}+1}}+{\frac {Cx+D}{x^{2}+4}}\\&{\frac {x^{3}+4x+3}{\left(x^{2}+4\right)}}=Ax+B+{\frac {Cx+D}{x^{2}+4}}\left(x^{2}+1\right)\\&x=i\\&\quad {\frac {-i+4i+3}{3}}=i+1=Ai+B\\&\qquad \qquad \qquad \qquad \qquad \quad A=1\qquad B=1\\&{\frac {x^{4}+4x^{2}+3x}{x^{4}+5x^{2}+4}}={\frac {x^{2}+x}{x^{2}+1}}+{\frac {Cx^{2}+Dx}{x^{2}+4}}\\&x\to \infty \\&\quad 1=1+C\qquad \qquad C=0\\&x=0\\&\quad {\frac {3}{4}}=1+{\frac {D}{4}}\qquad \quad D=-1\\&\int {\frac {x^{3}+4x+3}{\left(x^{2}+1\right)\left(x^{2}+4\right)}}dx\\&=\int {\frac {x+1}{x^{2}+1}}+{\frac {-1}{x^{2}+4}}dx\\&=\int {\frac {x}{x^{2}+1}}dx+\int {\frac {1}{x^{2}+1}}dx+\int {\frac {-1}{x^{2}+4}}dx\\&={\frac {\ln \left(x^{2}+1\right)}{2}}+\tan ^{-1}x-{\frac {1}{2}}\tan ^{-1}{\frac {x}{2}}\\&\\&2.\;\int {\frac {1}{x+x{\sqrt {x}}}}dx\\&=2\int {\frac {1}{{\sqrt {x}}+x}}d{\sqrt {x}}=2\int \left({\frac {A}{\sqrt {x}}}+{\frac {B}{{\sqrt {x}}+1}}\right)d{\sqrt {x}}\\&\quad A=1\quad B=-1\\&=2\int \left({\frac {1}{\sqrt {x}}}+{\frac {-1}{{\sqrt {x}}+1}}\right)d{\sqrt {x}}\\&=2\left(\int {\frac {1}{\sqrt {x}}}d{\sqrt {x}}-\int {\frac {1}{{\sqrt {x}}+1}}d{\sqrt {x}}\right)\\&=2\left(\ln {\sqrt {x}}-\ln \left({\sqrt {x}}+1\right)\right)\\&=2\ln {\frac {\sqrt {x}}{{\sqrt {x}}+1}}\end{aligned}}}
