Câu 1:
Ta có \(\int \frac{dx}{x^4+1}=\frac{1}{2}\int \left ( \frac{x^2+1}{x^4+1}-\frac{x^2-1}{x^4+1} \right )dx=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx+\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx\)
\(\frac{1}{2}\int \frac{d\left ( x-\frac{1}{x} \right )}{x^2+\frac{1}{x^2}}+\frac{1}{2}\int \frac{d\left ( x+\frac{1}{x} \right )}{x^2+\frac{1}{x^2}}=\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{(x-\frac{1}{x})^2+2}+\frac{1}{2}\int \frac{d(x+\frac{1}{2})}{(x+\frac{1}{x})^2-2}\)
Đặt \(x-\frac{1}{x}=a,x+\frac{1}{x}=b\Rightarrow A=\frac{1}{2}\int \frac{da}{a^2+2}+\frac{1}{2}\int \frac{db}{b^2-2}\)
Bằng cách đặt \(a=\sqrt{2}\tan u (-\frac{\pi}{2}< u<\frac{\pi}{2})\)
\(\Rightarrow \frac{1}{2}\int \frac{da}{a^2+2}=\frac{\sqrt{2}}{4}\tan^{-1}\left (\frac{a}{\sqrt{2}} \right)+c\)
\(\frac{1}{2}\int \frac{db}{b^2-2}=\frac{1}{4\sqrt{2}}\int \left (\frac{1}{b-\sqrt{2}}-\frac{1}{b+\sqrt{2}} \right)db\)\(=\frac{1}{4\sqrt{2}}\ln|\frac{b-\sqrt{2}}{b+\sqrt{2}}|+c\)
\(\Rightarrow A=\frac{1}{2\sqrt{2}}\tan^{-1} \left (\frac{x^2-1}{\sqrt{2}x} \right)-\frac{1}{4\sqrt{2}}\ln|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}|+c\)
Awn, chúc mừng năm mới!
Câu 2:
\(B=\int \frac{x^4+1}{x^6+1}=\int\frac{(x^2+1)^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx=\int\frac{x^2+1}{x^4-x^2+1}dx-2\int \frac{x^2dx}{(x^3)^2+1}\)
\(\int\frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}dx-\frac{2}{3}\int\frac{d(x^3)}{(x^3)^2+1}=\int\frac{d\left (x-\frac{1}{x} \right)}{\left (x-\frac{1}{x}\right)^2+1}-\frac{2}{3}\int\frac{d(x^3)}{(x^3)^2+1}\)
Đặt \(x-\frac{1}{x}=a, x^3=b\). Cần tính \(B=\int\frac{da}{a^2+1}-\frac{2}{3}\int\frac{db}{b^2+1}\)
Đến đây bài toán trở về dạng quen thuộc . Đặt \(a=\tan u, b=\tan v\)
\(\Rightarrow B=\tan ^{-1}\left (x-\frac{1}{x}\right)-\frac{2}{3}\tan^{-1}(x^3)+c\)
Câu 3:
\(C=\int\frac{x^3-x^2-4x-1}{x^3(x+1)}dx=\int \frac{dx}{x+1}-\int\frac{dx}{x(x+1)}-\int\frac{4dx}{x^3}+\int\frac{3}{x^3(x+1)}\)
Tính riêng lẻ từng phần :)
\(\int\frac{dx}{x+1}=\ln|x+1|;\int\frac{dx}{x(x+1)}=\int \left(\frac{1}{x}-\frac{1}{x+1}\right )dx=\ln |x|-\ln|x+1|\)
\(\int\frac{4dx}{x^3}=\frac{-2}{x^2}\)
\(\int\frac{3}{x^3(x+1)}=\int \frac{3}{x^2}\left ( \frac{1}{x}-\frac{1}{x+1} \right )dx=\int \frac{3dx}{x^3}-\int \frac{3dx}{x^2}+\int \frac{3dx}{x}-\int \frac{3dx}{x+1}=\frac{-3}{2x^2}+ \frac{3}{x}+3\ln|x|-3\ln|x+1|\)Suy ra \(C=2\ln|x|-\ln|x+1|+\frac{1}{2x^2}+\frac{3}{x}+c\)
Xong.
P/s: Đùa chứ bạn đào đâu ra toàn bài khoai @@
