\(y=\frac{1}{x^2+\sqrt{x}}\sqrt[]{}\sqrt{\frac{\left(^{ }\int^{ }_{ }\left(\right)\right)}{ }}\)=
Những câu hỏi liên quan
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 \(I_1=\int e^{t-2}dt-\frac{4\cdot e^{t-2}}{t}=e^{t-2}-\frac{4e^{t-2}}{t}+C\)
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3) Đặt \(t=\sqrt{1+\sqrt[3]{x^2}}\Rightarrow t^2-1=\sqrt[3]{x^2}\Leftrightarrow x^2=\left(t^2-1\right)^3\)
\(d\left(x^2\right)=d\left[\left(t^2-1\right)^3\right]\Leftrightarrow2x\cdot dx=6t\left(t^2-1\right)^2\cdot dt\)
\(I_3=\int\frac{3t\left(t^2-1\right)^2}{t}dt=3\int\left(t^4-2t^2+1\right)dt=...\)
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5) Đặt \(\frac{2+x}{2-x}=4t^3\Leftrightarrow4t^3=\frac{4}{2-x}-1\)
\(d\left(4t^3\right)=d\left(\frac{4}{2-x}-1\right)\Leftrightarrow3t^2dt=\frac{1}{\left(2-x\right)^2}dx\)
\(I_5=\int\frac{3t^2}{t\sqrt[3]{4}}dt=\frac{3}{\sqrt[3]{4}}\int tdt=...\)
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Xem thêm câu trả lời
1) int lnfrac{left(1+stext{inx}right)^{1+ctext{os}x}}{1+ctext{os}x}dx
2) intleft(xlnxright)^2dx
3) intfrac{3xcosx+2}{1+cot^2x}dx
4)intfrac{2}{ctext{os}2x-7}dx
5)intfrac{1+xleft(2lnx-1right)}{xleft(x+1right)^2}dx
6) intfrac{1-x^2}{left(1+x^2right)^2}dx
7)int e^xfrac{1+stext{inx}}{1+ctext{os}x}dx
8) int lnleft(frac{x+1}{x-1}right)dx
9)intfrac{xlnleft(1+xright)}{left(1+x^2right)^2}dx
10) intfrac{lnleft(x-1right)}{left(x-1right)^4}dx
11)intfrac{x^3lnx}{sqrt{x^2+1}}dx
12)intfrac{xe^x}{_{ }...
Đọc tiếp
1) \(\int ln\frac{\left(1+s\text{inx}\right)^{1+c\text{os}x}}{1+c\text{os}x}dx\)
2) \(\int\left(xlnx\right)^2dx\)
3) \(\int\frac{3xcosx+2}{1+cot^2x}dx\)
4)\(\int\frac{2}{c\text{os}2x-7}dx\)
5)\(\int\frac{1+x\left(2lnx-1\right)}{x\left(x+1\right)^2}dx\)
6) \(\int\frac{1-x^2}{\left(1+x^2\right)^2}dx\)
7)\(\int e^x\frac{1+s\text{inx}}{1+c\text{os}x}dx\)
8) \(\int ln\left(\frac{x+1}{x-1}\right)dx\)
9)\(\int\frac{xln\left(1+x\right)}{\left(1+x^2\right)^2}dx\)
10) \(\int\frac{ln\left(x-1\right)}{\left(x-1\right)^4}dx\)
11)\(\int\frac{x^3lnx}{\sqrt{x^2+1}}dx\)
12)\(\int\frac{xe^x}{_{ }\left(e^x+1\right)^2}dx\)
13) \(\int\frac{xln\left(x+\sqrt{1+x^2}\right)}{x+\sqrt{1+x^2}}dx\)
giúp mk đc con nào thì giúp nha
Câu 2)
Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2\frac{\ln x}{x}dx\\ v=\frac{x^3}{3}\end{matrix}\right.\Rightarrow I=\frac{x^3}{3}\ln ^2x-\frac{2}{3}\int x^2\ln xdx\)
Đặt \(\left\{\begin{matrix} k=\ln x\\ dt=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=\frac{dx}{x}\\ t=\frac{x^3}{3}\end{matrix}\right.\Rightarrow \int x^2\ln xdx=\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+c\)
Do đó \(I=\frac{x^3\ln^2x}{3}-\frac{2}{9}x^3\ln x+\frac{2}{27}x^3+c\)
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Câu 3:
\(I=\int\frac{2}{\cos 2x-7}dx=-\int\frac{2}{2\sin^2x+6}dx=-\int\frac{dx}{\sin^2x+3}\)
Đặt \(t=\tan\frac{x}{2}\Rightarrow \left\{\begin{matrix} \sin x=\frac{2t}{t^2+1}\\ dx=\frac{2dt}{t^2+1}\end{matrix}\right.\)
\(\Rightarrow I=-\int \frac{2dt}{(t^2+1)\left ( \frac{4t^2}{(t^2+1)^2}+3 \right )}=-\int\frac{2(t^2+1)dt}{3t^4+10t^2+3}=-\int \frac{2d\left ( t-\frac{1}{t} \right )}{3\left ( t-\frac{1}{t} \right )^2+16}=\int\frac{2dk}{3k^2+16}\)
Đặt \(k=\frac{4}{\sqrt{3}}\tan v\). Đến đây dễ dàng suy ra \(I=\frac{-1}{2\sqrt{3}}v+c\)
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Câu 6)
\(I=-\int \frac{\left ( 1-\frac{1}{x^2} \right )dx}{x^2+2+\frac{1}{x^2}}=-\int \frac{d\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2}=-\frac{1}{x+\frac{1}{x}}+c=-\frac{x}{x^2+1}+c\)
Câu 8)
\(I=\int \ln \left(\frac{x+1}{x-1}\right)dx=\int \ln (x+1)dx-\int \ln (x-1)dx\)
\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\) ta có:
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)
\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)
\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)
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Xem thêm câu trả lời
rút gọn:a)left(frac{1}{2+2sqrt{x}}+frac{1}{2-2sqrt{x}}-frac{x^2+1}{1-x^2}right)timesleft(1+frac{1}{x}right)b)left(frac{2sqrt{xy}}{x-y}+frac{sqrt{x}-sqrt{y}}{2sqrt{x}+sqrt{y}}right)timesfrac{2sqrt{x}}{sqrt{x}+sqrt{y}}+frac{sqrt{y}}{sqrt{y}-sqrt{x}}c)left(frac{x-1}{sqrt{x}-1}+frac{xsqrt{x}-1}{1-x}right)divfrac{left(sqrt{x}-1right)^2+sqrt{x}}{sqrt{x}+1}
Đọc tiếp
rút gọn:
a)\(\left(\frac{1}{2+2\sqrt{x}}+\frac{1}{2-2\sqrt{x}}-\frac{x^2+1}{1-x^2}\right)\times\left(1+\frac{1}{x}\right)\)
b)\(\left(\frac{2\sqrt{xy}}{x-y}+\frac{\sqrt{x}-\sqrt{y}}{2\sqrt{x}+\sqrt{y}}\right)\times\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)
c)\(\left(\frac{x-1}{\sqrt{x}-1}+\frac{x\sqrt{x}-1}{1-x}\right)\div\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}+1}\)
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
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Pleft(frac{sqrt{x}-2}{x-1}-frac{sqrt{x}+2}{x+2sqrt{x}+1}right)frac{left(1-xright)^2}{2}Pleft(frac{sqrt{x}-2}{left(sqrt{x}-1right)left(sqrt{x}+1right)}-frac{sqrt{x}+2}{left(sqrt{x}+1right)^2}right)frac{left(1-xright)^2}{2}Pleft(frac{left(sqrt{x}-2right)left(sqrt{x}+1right)-left(sqrt{x}+2right)left(sqrt{x}-1right)}{left(x-1right)left(sqrt{x}+1right)}right)frac{left(x-1right)^2}{2}Pleft(frac{left(x-sqrt{x}-2right)-left(x+sqrt{x}-2right)}{left(x-1right)left(sqrt{x}+1right)}right)frac{left(x-1right)^...
Đọc tiếp
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right)\frac{\left(x-1\right)^2}{2}\)
\(P=\left(\frac{\left(x-\sqrt{x}-2\right)-\left(x+\sqrt{x}-2\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right)\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{\sqrt{x}\left(x-1\right)}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}\)
Rút gọn:
\(A=\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right]\)
\(x=\sqrt{2-\sqrt{3}};y=\sqrt{2+\sqrt{3}}\)
Biết \(0< x\le y\)và \(\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)+2\left(x+2y\right)}\right)+\left(\frac{y}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}\right)=\frac{5}{3}\)
Tính \(\frac{x}{y}\)
\(\int^{\left(x-1\right)^2+y^2=\sqrt[3]{x\left(2x+1\right)}}_{3x^2-x+\frac{1}{2}=y\sqrt{x^2+x}}\) giải hpt.............
frac{x+1}{2left(x-1right)}+frac{2}{2left(sqrt{x}+1right)}+frac{sqrt{x}}{sqrt{x}left(sqrt{x}-1right)}.frac{left(x+1right).sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2sqrt{x}left(sqrt{x}-1right)}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2sqrt{x}left(sqrt{x}+1right)}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}.frac{xsqrt{x}+sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2x-2sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2x+2...
Đọc tiếp
\(=\frac{x+1}{2\left(x-1\right)}+\frac{2}{2\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\)
=\(\frac{\left(x+1\right).\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x-2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x+2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+4x+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(x+4\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
LƯU Ý: CAP NÀY CHỈ LÀ CAP NHÁP
1/Rút gọn
Afrac{left(sqrt{x}+1right)left(x-sqrt{xy}right)left(sqrt{x}+sqrt{y}right)}{left(x-yright)left(sqrt{x^3+x}right)}(x0; y0; x#y)
B left(frac{1}{sqrt{x}+1}-frac{1}{x+sqrt{x}}right):frac{x-sqrt{x}+1}{xsqrt{x}+1}( x0)
Cleft(frac{x+1}{sqrt{x}}+2right).frac{sqrt{x}}{left(sqrt{x}+1right)left(xsqrt{x}+1right)}(x0)
Dleft(frac{xsqrt{x}-1}{sqrt{x}-1}+sqrt{x}right):left(x-1right)-frac{2}{sqrt{x}-1}(x0; x#1)
giúp em với ạ em đang cần gấp ạ
Đọc tiếp
1/Rút gọn
A=\(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{xy}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\left(\sqrt{x^3+x}\right)}\)(x>0; y>0; x#y)
B= \(\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)( x>0)
C=\(\left(\frac{x+1}{\sqrt{x}}+2\right).\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x\sqrt{x}+1\right)}\)(x>0)
D=\(\left(\frac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right):\left(x-1\right)-\frac{2}{\sqrt{x}-1}\)(x>=0; x#1)
giúp em với ạ em đang cần gấp ạ