giải giúp mình 2 con nguyên hàm này vơi
\(A=\int \frac{x\sin x+\cos x}{x^2-\cos ^2x}dx\)
\(B=\int \frac{\ln x-1}{x^2-\ln ^2x}dx\)
giải giúp mình 2 con nguyên hàm này vơi
\(A=\int \frac{x\sin x+\cos x}{x^2-\cos ^2x}dx\)
\(B=\int \frac{\ln x-1}{x^2-\ln ^2x}dx\)
Lời giải:
Ta có:
\(A=\int \frac{x\sin x+\cos x}{x^2-\cos ^2x}dx=\int \frac{(\cos x-x)+x(\sin x+1)}{x^2-\cos ^2x}dx\)
\(=-\int \frac{dx}{\cos x+x}+\int \frac{x(\sin x+1)}{x^2-\cos ^2x}dx=-\int \frac{dx}{x+\cos x}+\frac{1}{2}\int (\sin x+1)\left(\frac{1}{x-\cos x}+\frac{1}{x+\cos x}\right)dx\)
\(=-\int \frac{dx}{x+\cos x}+\frac{1}{2}\int (\sin x+1)\frac{dx}{x-\cos x}+\frac{1}{2}\int (\sin x-1)\frac{dx}{x+\cos x}+\int \frac{dx}{x+\cos x}\)
\(=\frac{1}{2}\int (\sin x+1)\frac{dx}{x-\cos x}+\frac{1}{2}\int (\sin x-1)\frac{dx}{x+\cos x}\)
\(=\frac{1}{2}\int \frac{d(x-\cos x)}{x-\cos x}+\frac{1}{2}\int \frac{-d(x+\cos x)}{x+\cos x}\)
\(=\frac{1}{2}\ln |x-\cos x|-\frac{1}{2}\ln |x+\cos x|+c\)
Xét biểu thức $B$
\(B=\int \frac{\ln x-1}{x^2-\ln ^2x}dx=\int \frac{(\ln x-x)+(x-1)}{x^2-\ln ^2x}dx\)
\(=-\int \frac{dx}{x+\ln x}+\int \frac{x-1}{x^2-\ln ^2x}dx=-\int \frac{dx}{x+\ln x}+\frac{1}{2}\int \frac{(x-1)}{x}\left(\frac{1}{x-\ln x}+\frac{1}{x+\ln x}\right)dx\)
\(=-\int \frac{dx}{x+\ln x}+\frac{1}{2}\int \frac{1}{x-\ln x}.\frac{x-1}{x}dx+\frac{1}{2}\int \frac{1}{x+\ln x}.\frac{x-1}{x}dx\)
\(=-\int \frac{dx}{x+\ln x}+\frac{1}{2}\int \frac{1}{x-\ln x}.\frac{x-1}{x}dx-\frac{1}{2}\int \frac{1}{x+\ln x}.\frac{1+x}{x}dx+\int \frac{dx}{x+\ln x}\)
\(=\frac{1}{2}\int \frac{1}{x-\ln x}.\frac{x-1}{x}dx-\frac{1}{2}\int \frac{1}{x+\ln x}.\frac{1+x}{x}dx\)
\(=\frac{1}{2}\int \frac{d(x-\ln x)}{x-\ln x}-\frac{1}{2}\int \frac{d(x+\ln x)}{x+\ln x}\)
\(=\frac{1}{2}\ln |x-\ln x|-\frac{1}{2}\ln |x+\ln x|+c\)
\(\int xsin\sqrt{x}\)dx
Lời giải:
Đặt \(x=t^2\Rightarrow I=\int t^2\sin td(t^2)=2\int t^3\sin tdt\)
Đặt \(\left\{\begin{matrix} u_1=t^3\\ dv_1=\sin tdt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du_1=3t^2dt\\ v_1=-\cos t\end{matrix}\right.\Rightarrow I=-t^3\cos t+3\int t^2\cos tdt\)
Tiếp tục
Đặt \(\left\{\begin{matrix} u_2=t^2\\ dv_2=\cos tdt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du_2=2tdt\\ v_2=\sin t\end{matrix}\right.\Rightarrow I=-t^3\cos t+3t^2\sin t-6\int t\sin tdt\)
Tiếp tục nguyên hàm từng phần cho \(\int t\sin tdt\)
\(\Rightarrow I=-t^3\cos t +3t^2\sin t+6t\cos t-6\sin t+c\)
Giúp em với : Tính nguyên hàm sau
Lời giải:
Phân tích:
\(I=\int\frac{x^8}{x^3+x+2}dx=\underbrace{\int (x^5-x^3-2x^2+x+4)dx}_{A}+\underbrace{\int \frac{3x^2-6x-8}{x^3+x+2}dx}_{B}\)
Có \(A=\frac{x^6}{6}-\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+4x+c(1)\)
\(B=\int\frac{3(x^2-x+2)-3(x+1)-11}{(x^2-x+2)(x+1)}dx\) \(=3\int\frac{dx}{x+1}-3\int\frac{dx}{x^2-x+2}-\int\frac{11dx}{x^3+x+2}\)
Đối với \(\int\frac{dx}{x^2-x+2}=\int\frac{dx}{(x-\frac{1}{2})^2+\frac{7}{4}}\) ta đặt \(x-\frac{1}{2}=\frac{\sqrt{7}}{2}\tan t\)
\(\Rightarrow \int\frac{dx}{x^2-x+2}=\frac{2\sqrt{7}}{7}\tan ^-1\left(\frac{2x-1}{\sqrt{7}}\right)+c\)
Đối với
\(\int\frac{dx}{x^3+x+2}=\int\frac{d(x^3+x+2)}{x^3+x+2}-\int\frac{3x^2dx}{x^3+x+2}=\ln|x^3+x+2|-\int\frac{3dx}{x+1}-\int\frac{3dx}{x^2-x+2}+\int\frac{9dx}{x^3+x+2}\)
\(\Rightarrow -8\int\frac{dx}{x^3+x+2}=\ln|x^3+x+2|-3\ln|x+1|-\frac{6\sqrt{7}}{7}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}}\right)\)
\(\Rightarrow \int\frac{dx}{x^3+x+2}=\frac{-\ln|x^2-x+2|}{8}+\frac{\ln|x+1|}{4}+\frac{3\sqrt{7}}{28}\tan^-1\left(\frac{2x-1}{\sqrt{7}}\right)\)
Vậy \(B=\frac{\ln|x+1|}{4}+\frac{11\ln|x^2-x+2|}{8}-\frac{57\sqrt{7}}{28}\tan^-1\left(\frac{2x-1}{\sqrt{7}}\right)+c(2)\)
Từ \((1),(2)\Rightarrow I=\frac{x^6}{6}-\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+4x+\frac{\ln|x+1|}{4}+\frac{11\ln|x^2-x+2|}{8}-\frac{57\sqrt{7}}{28}\tan^-1\left(\frac{2x-1}{\sqrt{7}}\right)+c\)
Lớp 12 là lớp lớn nhất rồi a còn xưng em em làm j???
\(\int\frac{x^2+3x+3}{x^3-3x+2}dx\)
Lời giải:
\(\int\frac{x^2+3x+3}{x^3-3x+2}dx=\int\frac{(x+2)(x-1)+2(x+2)+1}{(x-1)^2(x+2)}dx=\int\frac{7dx}{3(x-1)^2}+\int\frac{8dx}{9(x-1)}+\int\frac{dx}{9(x+2)}\)\(=\frac{-7}{3(x-1)}+\frac{8}{9}\ln|x-1|+\frac{1}{9}\ln|x+2|+c\)
\(\int\frac{x}{\left(1+2x\right)^3}dx\)
\(\int\frac{1-x^2}{x+x^3}dx\)
1)Đặt \(1+2x=t\Leftrightarrow x=\frac{t-1}{2}; dx=\frac{dt}{2}.\)
\(I_1=\frac{1}{4}\int\frac{t-1}{t^3}dt=\frac{1}{4}\int\left(\frac{1}{t^2}-\frac{1}{t^3}\right)dt=...\)
2) \(\int\frac{1-x^2}{x+x^3}dx=\int\left(\frac{1}{x}-\frac{2x}{1+x^2}\right)dx=\int\frac{dx}{x}-\int\frac{d\left(1+x^2\right)}{1+x^2}=...\)
\(\int\frac{1}{x^4+1}dx\)
\(\int\frac{x^4+1}{x^6+1}dx\)
\(\int\frac{x^3-x^2-4x-1}{x^4+x^3}dx\)
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 @@
\(\int\sqrt{e^x-1}dx\)
\(\int\frac{\sqrt{1+x^2}}{x^4}dx\)
Câu 1:Gọi biểu thức là $A$. Đặt \(\sqrt{e^x-1}=t\)
\(\Rightarrow e^x=t^2+1\Rightarrow d(e^x)=d(t^2+1)=2tdt=e^xdx=(t^2+1)dx\)
\(\Rightarrow \int \frac{2t^2}{t^2+1}dt=\int \left (2-\frac{2}{t^2+1} \right)dt\)
Đặt \(t=\tan m\Rightarrow dt=\frac{dm}{\cos^2 m}\Rightarrow \int \frac{2dt}{t^2+1}=\int 2dm=2m\)
\(\Rightarrow A=2t-2m+c=2\sqrt{e^x-1}-2\tan ^{-1} (\sqrt{e^x-1})+c\)
Câu 2: Đặt \(x=\tan t\Rightarrow dx=\frac{dt}{\cos^2 t}, x^2+1=\frac{1}{\cos^2 t}\) với \(\frac{-\pi}{2} < t< \frac{\pi}{2}\)
Gọi biểu thức là $B$. Ta có
\(B=\int \frac{\cos t dt}{\sin ^4t}=\int \frac{d(\sin t)}{\sin^4 t}=\frac{-\sin ^{-3} t}{3}+c\) \(=-\frac{\sqrt{(x^2+1)^3}}{3x^3}+c\)
1) \(\int\left(\frac{lnx}{2+lnx}\right)^2\)
2) \(\int\frac{dx}{\left(x+3\right)^3\left(x+5\right)^5}\)
3) \(\int\frac{xdx}{\sqrt{1+\sqrt[3]{x^2}}}\)
4) \(\int\frac{dx}{x^3.\sqrt[3]{2-x^3}}\)
5)\(\int\sqrt[3]{\frac{2-x}{2+x}}.\frac{1}{\left(2-x\right)^2}dx\)
1) Đặt \(2+lnx=t\Leftrightarrow x=e^{t-2}\Rightarrow dx=e^{t-2}dt\)
\(I_1=\int\left(\frac{t-2}{t}\right)^2\cdot e^{t-2}\cdot dt=\int\left(1-\frac{4}{t}+\frac{4}{t^2}\right)e^{t-2}dt\\ =\int e^{t-2}dt-4\int\frac{e^{t-2}}{t}dt+4\int\frac{e^{t-2}}{t^2}dt\)
Có:
\(4\int\frac{e^{t-2}}{t^2}dt=-4\int e^{t-2}\cdot d\left(\frac{1}{t}\right)=-\frac{4\cdot e^{t-2}}{t}+4\int\frac{e^{t-2}}{t}dt\\ \Leftrightarrow4\int\frac{e^{t-2}}{t^2}dt-4\int\frac{e^{t-2}}{t^{ }}dt=-\frac{4\cdot e^{t-2}}{t}\)
Vậy \(I_1=\int e^{t-2}dt-\frac{4\cdot e^{t-2}}{t}=e^{t-2}-\frac{4e^{t-2}}{t}+C\)
3) Đặt \(t=\sqrt{1+\sqrt[3]{x^2}}\Rightarrow t^2-1=\sqrt[3]{x^2}\Leftrightarrow x^2=\left(t^2-1\right)^3\)
\(d\left(x^2\right)=d\left[\left(t^2-1\right)^3\right]\Leftrightarrow2x\cdot dx=6t\left(t^2-1\right)^2\cdot dt\)
\(I_3=\int\frac{3t\left(t^2-1\right)^2}{t}dt=3\int\left(t^4-2t^2+1\right)dt=...\)
5) Đặt \(\frac{2+x}{2-x}=4t^3\Leftrightarrow4t^3=\frac{4}{2-x}-1\)
\(d\left(4t^3\right)=d\left(\frac{4}{2-x}-1\right)\Leftrightarrow3t^2dt=\frac{1}{\left(2-x\right)^2}dx\)
\(I_5=\int\frac{3t^2}{t\sqrt[3]{4}}dt=\frac{3}{\sqrt[3]{4}}\int tdt=...\)
\(\int\frac{dx}{2sinx+5cosx+3}\)
Đặt x=2t, dx=2dt
\(2sinx+5cosx+3=2sin2t+5cos2t+3\\ =4sint\cdot cost+5\left(cos^2t-sin^2t\right)+3\left(sin^2t+cos^2t\right)\\ =-2sin^2t+4sint\cdot cost+8cos^2t\)
Ta có:
\(I=\int\frac{2dt}{-2sin^2t+4sint\cdot cost+8cos^2t}\\ =\int\frac{\frac{dt}{cos^2t}}{-tan^2t+2tant+4}=\int\frac{d\left(tant\right)}{-tan^2t+2tant+4}\\ =\int\frac{-d\left(tant\right)}{\left(tant-1+\sqrt{5}\right)\left(tant-1-\sqrt{5}\right)}\\ =\frac{1}{2\sqrt{5}}\int\left(\frac{1}{tant-1+\sqrt{5}}-\frac{1}{tant-1-\sqrt{5}}\right)dt\)
\(=\frac{1}{2\sqrt{5}}ln\left|\frac{tant-1+\sqrt{5}}{tant-1+\sqrt{5}}\right|+C\)
....