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ta có
\(\int\frac{2x+1}{\left(x^2+x+1\right)^3}dx=\int\frac{d\left(x^2+x+1\right)}{\left(x^2+x+1\right)^3}=-\frac{1}{2\left(x^2+x+1\right)^2}+C\)
TA có
\(\int\frac{x+2}{x\left(x-3\right)}dx=\int\frac{x-3+5}{x\left(x-3\right)}dx=\int\left(\frac{1}{x}+\frac{5}{x\left(x-3\right)}\right)dx=\int\frac{1}{x}dx+5\int\frac{1}{x\left(x-3\right)}dx\)
=\(\int\frac{1}{x}dx+\frac{5}{3}\int\left(\frac{1}{x-3}-\frac{1}{x}\right)dx=-\frac{2}{3}\int\frac{1}{x}dx+\frac{5}{3}\int\frac{1}{x-3}dx=\frac{-2}{3}ln\left|x\right|+\frac{5}{3}ln\left|x-3\right|+C\)
Ta có
\(\int\frac{dx}{x^2-4x+3}=\int\frac{dx}{\left(x-1\right)\left(x-3\right)}=\frac{1}{2}\int\left(\frac{1}{x-3}-\frac{1}{x-1}\right)dx=\frac{1}{2}\left(ln\left|x-3\right|-ln\left|x-1\right|\right)+C=\frac{1}{2}ln\left|\frac{x-3}{x-1}\right|+C\)
ta có
\(\int\frac{dx}{\sqrt{x^2+1}+\sqrt{2-x^2}}=\int\frac{\sqrt{x^2+1}+\sqrt{2-x^2}}{\left(\sqrt{x^2+1}+\sqrt{1-x^2}\right)\left(\sqrt{x^2+1}-\sqrt{2-x^2}\right)}dx=\int\frac{\sqrt{x^2+1}+\sqrt{2-x^2}}{3x^2}dx=\int\frac{\sqrt{x^2+1}}{3x^2}+\int\frac{\sqrt{2-x^2}dx}{3x^2}\)=\(\frac{1}{3}\left(I_1+I_2\right)\)
Tính \(I_1=\int\frac{\sqrt{1+x^2}}{x^2}dx\)
\(tant=x\Rightarrow dx=\frac{1}{cos^2x}dx\)
ta có
\(\int\frac{\sqrt{1+tan^2t}}{cos^2t.tan^2t}dt=\int\frac{\frac{1}{cosx}}{sin^2t}dx=\int\frac{d\left(sint\right)}{sin^2t\left(1-sin^2t\right)}=\int\frac{dy}{y^2\left(1-y^2\right)}\)
làm tương tự câu trên ta tính đc \(I_1,I_2\) TA TÍNH ĐC I
a)
\(\frac{1}{x^2+x+1}dx=\frac{1}{\left(x-\frac{1}{4}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx\)
Đặt
\(\left(x-\frac{1}{4}\right)=\frac{\sqrt{3}}{2}tant\) => dx=\(\frac{\sqrt{3}}{2}\left(1+tan^2t\right)dt\) =>\(\frac{1}{x^2+x+1}dx=\frac{1}{\frac{3}{4}\left(1+tan^2t\right)+\frac{3}{4}}\left(1+tan^2t\right)dt=\frac{3}{4}dt=\frac{3}{4}t+C\)
Với \(\left(x-\frac{1}{4}\right)=\frac{\sqrt{3}}{2}tant=>t=\left(\frac{2\sqrt{3}}{4x-1}\right)\)
Câu b nhá :
\(\frac{1}{x^2+2x+2}dx=\frac{1}{\left(x+1\right)^2+\left(\sqrt{2^2}\right)}dx\)
Đặt
\(x+1=\sqrt{2}tant=>dx=\sqrt{2}\left(1+tan^2t\right)dt\)
=> \(\frac{1}{x^2+2x+3}dx=\frac{1}{2\left(tan^2t+1\right)}.\left(1+tan^2t\right)dt=\frac{1}{2}dt=\frac{1}{2}t+C\)
Với
\(x+1=\sqrt{2}tant=>tant=\frac{x+1}{\sqrt{2}}<=>t=arctan\left(\frac{x+1}{\sqrt{2}}\right)\)