Question

1.Calculati:
a) ∫$\frac{x^2+4x+5} {x^2+4x+3}$ dx=
b)∫$\frac{1}{x^2+2x}$ dx=
c)∫(1-$\frac{1}{x^2}$)lnx dx=

Answer 1

a) $\int\frac{x^2+4x+5}{x^2+4x+3}dx=\int\frac{x^2+4x+3+2}{x^2+4x+3}dx=\int\frac{x^2+4x+3}{x^2+4x+3}+\frac{2}{x^2+4x+3}dx=\int1dx+2\int\frac{1}{(x+2)^2-1}dx=\\=x+2\frac12\ln\big|\frac{x+2-1}{x+2+1}\big|+\mathcal{C}=x+\ln\big|\frac{x+1}{x+3}\big|+\mathcal{C}$

$b)\,\int\frac{1}{x^2+2x}dx=\int\frac{1}{x(x+2)}dx=I_1\\\text{Descompunem }\frac1{x(x+2)}\text{ \^in frac\c tii simple:}\\\\\frac1{x(x+2)}=\frac{A}x+\frac{B}{x+2}\\1=A(x+2)+Bx\\1=Ax+2A+Bx\\1=(A+B)x+2A\\\left \{ {{2A=1} \atop {A+B=0}} \right. \leftrightarrow\left \{ {{A=\frac12} \atop {B=\frac{-1}2}} \right. \\\\\frac1{x(x+2)}=\frac12\big(\frac1x-\frac{1}{x+2}\big)\\I_1=\int\frac12\big(\frac1x-\frac1{x+2}\big)dx=\frac12\big(\int\frac1xdx-\int\frac1{x+2}dx\big)=\frac12(\ln|x|-\ln|x+2|)+\mathcal{C}$

c) $I=\int\big(1-\frac1{x^2}\big)\ln xdx=\int\ln xdx+\int\frac{-1}{x^2}\ln xdx=I_1+I_2\\I_1=\int\ln xdx=x\ln x-\int x\cdot (\ln x)'dx=x\ln x-\int x\cdot \frac1x dx=x\ln x-x+\mathcal{C}\\I_2=\int\frac{-1}{x^2}\ln xdx\\\text{Substitu\c tie }\frac{1}{x}=t\rightarrow x=\frac1t\\-\frac1{x^2}dx=dt\\I_2=\int\ln \frac1t dt=t\ln\frac1t-\int t\cdot(\ln\frac1t)'dt=t\ln\frac1t-\int t\cdot \frac{-\frac1{t^2}}{\frac1t}dt=t\ln\frac1t-\int t\cdot\frac{-1}{t^2}\cdot tdt=\\=t\ln\frac1t+\int1dt=t\ln\frac1t+t+\mathcal{C}=\frac1x\ln x+\frac1x+\mathcal{C}\\I=x\ln x-x+\frac1x\ln x+\frac1x+\mathcal{C}=-x+\frac1x+(x+\frac1x)\ln x+\mathcal{C}$