Biết \(\int f(x) \mathrm{d}x\)=2xe2x+1 + C .Tìm \(\int f(2x) \mathrm{d}x\)
Tìm các nguyên hàm sau:
a) \(\int (3x^2-2x-4)dx \)
b) \(\int(\sin3x-\cos4x)dx \)
c) \(\int(e^{-3x}-4^x)dx \)
d) \(\int\ln(x)dx \)
e) \(\int(x.e^x)dx \)
f) \(\int(x+1).\sin(x)dx \)
g) \(\int x.\ln(x)dx \)
\(\int\left(3x^2-2x-4\right)dx=x^3-x^2-4x+C\)
\(\int\left(sin3x-cos4x\right)dx=-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+C\)
\(\int\left(e^{-3x}-4^x\right)dx=-\dfrac{1}{3}e^{-3x}-\dfrac{4^x}{ln4}+C\)
d. \(I=\int lnxdx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x\end{matrix}\right.\)
\(\Rightarrow u=x.lnx-\int dx=x.lnx-x+C\)
e. Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=x.e^x-\int e^xdx=x.e^x-e^x+C\)
f.
Đặt \(\left\{{}\begin{matrix}u=x+1\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cosx\end{matrix}\right.\)
\(\Rightarrow I=-\left(x+1\right)cosx+\int cosxdx=-\left(x+1\right)cosx+sinx+C\)
g.
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)
\(a)x+2x+3x+4x+...+100x=-213 b)\frac12x-\frac13=\frac14x-\frac16 c)3.(x-2)+2.(x-1)=10 d)\frac{\mathrm x+1}{\mathrm 3}=\frac{\mathrm x-2}{\mathrm 4} \)
a) \(\int sin2x.cosxdx\)
b) \(\int tanxdx\)
c) \(\int\dfrac{sinx}{1+3cosx}dx\)
d) \(\int sin^3xdx\)
e) \(\int sin^2xdx\)
f) \(\int cos^23x\)
g) \(f\left(x\right)=\dfrac{1}{sin^2x.cos^2x}\)
h) \(f\left(x\right)=\dfrac{cos2x}{sin^2x.cos^2x}\)
i) \(\int2sin3x.cos2xdx\)
j) \(\int e^x\left(2+\dfrac{e^{-x}}{cos^2x}\right)dx\)
\(a,\int sin2x.cosxdx=\int\dfrac{1}{2}\left[sin3x+sinx\right]dx=\dfrac{1}{2}\int sin3xdx+\dfrac{1}{2}\int sinxdx=\dfrac{-1}{6}cos3x-\dfrac{1}{2}cosx\)
phần a bạn thêm +C vào đáp án nhé
\(i,\int2sinx3x.cos2xdx=2\int\dfrac{1}{2}\left(sin5x+sinx\right)dx=\int sin5xdx+\int sinxdx=-\dfrac{1}{5}cos5x-cosx+C\)
\(g,\int\dfrac{1}{sin^2x.cos^2x}=\int\dfrac{sin^2x+cos^2x}{sin^2x.cos^2x}=\int\dfrac{1}{cos^2x}dx+\int\dfrac{1}{sin^2x}dx=tanx-cotx+C\)
C2. Xác định các chất tương ứng với các chữ cái \( \mathrm{A}, \mathrm{B}, \mathrm{D}, \mathrm{E}, \mathrm{F}, \mathrm{G}, \mathrm{H}, \mathrm{I}, \mathrm{K}, \mathrm{L} \), \( \mathrm{M}, \mathrm{N}, \mathrm{P}, \mathrm{Q} \) và hoàn thành các phương trình phản ứng.
(1) \( \mathrm{FeS}_{2}+\mathrm{O}_{2} \rightarrow \mathrm{A}+\mathrm{B} \uparrow \)
(5) \( \mathrm{H}+\mathrm{M} \rightarrow \mathrm{N} \)
(2) \( \mathrm{B}+\mathrm{D} \rightarrow \mathrm{E} \downarrow+\mathrm{F} \)
(6) \( \mathrm{N}+\mathrm{I} \rightarrow \mathrm{P} \downarrow+\mathrm{L} \) (vàng)
(7) \( \mathrm{P}+\mathrm{F}+\mathrm{Q} \rightarrow \mathrm{K} \)
(3) \( \mathrm{A}+\mathrm{G} \rightarrow \mathrm{H}+\mathrm{F} \)
(8) \( \mathrm{K}+\mathrm{G}+\mathrm{M} \rightarrow \mathrm{N}+\mathrm{F} \)
(4) \( \mathrm{H}+\mathrm{I} \rightarrow \mathrm{K} \downarrow+\mathrm{L} \)
Tìm x biết:
\(a) x+2x+3x+4x+...+100x=-213\)
\(b) \frac12 x-\frac13=\frac14-4\frac16\)
\(c)3(x-2)+2(x-1)=10\)
\(d)\frac{\mathrm x+1}{\mathrm3}=\frac{\mathrm x-2}{\mathrm4}\)
a) \(x+2x+3x+...+100x=-213\)
\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)
\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)
\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)
d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
c) \(3\left(x-2\right)+2\left(x-1\right)=10\)
\(\Rightarrow3x-6+2x-2=10\)
\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)
a) \(x+2x+3x+4x+...+100x=-213\)
\(x.\left(1+2+3+4+...+100\right)=-213\)
\(x.5050=-213\)
\(x=-\frac{213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)
\(\frac{1}{2}x=-\frac{43}{12}\)
\(x=\frac{-43}{6}\)
c) \(3\left(x-2\right)+2\left(x-1\right)=10\)
\(3x-6+2x-2=10\)
\(3x+2x-\left(6+2\right)=10\)
\(5x-8=10\)
\(5x=18\)
\(x=\frac{18}{5}\)
d) \(\frac{x+1}{3}=\frac{x-2}{4}\)
\(\Rightarrow4.\left(x+1\right)=3\left(x-2\right)\)
\(4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\)
\(x=-10\)
cho \(\int f\left(4x\right)dx\) = x2+3x+C. Mệnh đề nào sau đây đúng?
A. \(\int f\left(x+2\right)dx\) =x2+7x+C
B.\(\int f\left(x+2\right)dx\) =\(\frac{x^2}{2}\)+4x+C
C.\(\int f\left(x+2\right)dx\)=\(\frac{x^2}{4}\)+2x+C
D.\(\int f\left(x+2\right)dx\)=\(\frac{x^2}{4}\)+4x+C
Giúp mình bài này với, cám ơn mọi người nhiều
\(\int f\left(4x\right)dx=\frac{1}{4}\int f\left(4x\right)d\left(4x\right)=\frac{1}{16}\left(4x\right)^2+\frac{3}{4}\left(4x\right)+C\)
\(\Rightarrow\int f\left(4x\right)d\left(4x\right)=\frac{1}{4}\left(4x\right)^2+3.\left(4x\right)+C\)
\(\Rightarrow\int f\left(x+2\right)dx=\int f\left(x+2\right)d\left(x+2\right)=\frac{1}{4}\left(x+2\right)^2+3\left(x+2\right)+C\)
\(=\frac{1}{4}x^2+4x+C\)
Cho biểu thức P=$
\frac{{x}^{2}}{\left({{x}\mathrm{{+}}{y}}\right)\left({{1}\mathrm{{-}}{y}}\right)}\mathrm{{-}}\frac{{y}^{2}}{\left({{x}\mathrm{{+}}{y}}\right)\left({{1}\mathrm{{+}}{x}}\right)}\mathrm{{-}}\frac{{x}^{2}{y}^{2}}{{\mathrm{(}}{x}\mathrm{{+}}{y}{\mathrm{)(}}{1}\mathrm{{-}}{y}{\mathrm{)}}}
$
a, Rút gọn P
b, Tìm các cặp số (x;y) thuộc Z sao cho giá trị của P=3
Nếu \(\int\limits^2_1\) f(x) dx = -2 và \(\int\limits^3_2\) f(x) dx =1 thì \(\int\limits^3_1\) f(x) dx bằng
A. -3
B. -1
C. 1
D. 3
\(\int\limits^3_1f\left(x\right)dx=-2+1=-1\)
Mọi người ơi , giúp e tính tích phân bất định với ạ ! Cảm ơn m.n ạ !
a.\(\int\frac{x+6}{\sqrt{x^2-2x+10}}dx\)
b.\(\int\frac{x}{\sqrt{3-2x-x^2}}dx\)
c.\(\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx\)
d,\(\int\frac{dx}{1+tanx}\)
e.\(\int tan^3xdx\)
f. \(\int cos^3xdx\)
g. \(\int sin^2x.cos^3xdx\)
h. \(\int sinx.cos2xdx\)
i. \(\int\frac{sin2x}{1+cos^2x}dx\)
a.
\(I=\int\frac{\frac{1}{2}\left(2x-2\right)+7}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx+7\int\frac{1}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}I_1+7I_2\)
Xét \(I_1=\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx=\int\frac{d\left(x^2-2x+10\right)}{\sqrt{x^2-2x+10}}=2\sqrt{x^2-2x+10}+C_1\)
Xét \(I_2=\int\frac{dx}{\sqrt{x^2-2x+10}}=\int\frac{dx}{\sqrt{\left(x-1\right)^2+9}}\)
Đặt
\(u=x-1+\sqrt{\left(x-1\right)^2+10}\Rightarrow du=\left(1+\frac{\left(x-1\right)}{\sqrt{\left(x-1\right)^2+10}}\right)dx=\frac{x-1+\sqrt{\left(x-1\right)^2+10}}{\sqrt{\left(x-1\right)^2+10}}dx\)
\(\Rightarrow du=\frac{u}{\sqrt{\left(x-1\right)^2+10}}dx\Rightarrow\frac{dx}{\sqrt{\left(x-1\right)^2+10}}=\frac{du}{u}\)
\(\Rightarrow I_2=\int\frac{du}{u}=ln\left|u\right|+C_2=ln\left|x-1+\sqrt{x^2-2x+10}\right|+C_2\)
\(\Rightarrow I=\sqrt{x^2-2x+10}+7ln\left|x-1+\sqrt{x^2-2x+10}\right|+C\)
2.
\(I=\int\frac{\frac{1}{2}\left(2x+2\right)-1}{\sqrt{3-2x-x^2}}dx=\frac{1}{2}\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx-\int\frac{1}{\sqrt{3-2x-x^2}}dx=\frac{1}{2}I_1-I_2\)
Xét \(I_1=\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx=-\int\frac{d\left(3-2x-x^2\right)}{\sqrt{3-2x-x^2}}=-2\sqrt{3-2x-x^2}+C_1\)
Xét \(I_2=\int\frac{1}{\sqrt{3-2x-x^2}}dx=\int\frac{1}{\sqrt{4-\left(x+1\right)^2}}dx\)
Đặt \(x+1=2sinu\Rightarrow dx=2cosu.du\)
\(\Rightarrow I_2=\int\frac{2cosu.du}{2.cosu}=\int du=u+C_2=arcsin\left(\frac{x+1}{2}\right)+C_2\)
\(\Rightarrow I=-\sqrt{3-2x-x^2}-arcsin\left(\frac{x+1}{2}\right)+C\)
c/
\(I=\int\frac{1-\sqrt{x}}{\sqrt{1-x}}dx\)
Đặt \(\sqrt{x}=sint\Rightarrow x=sin^2t\Rightarrow dx=2sint.cost.dt\)
\(\Rightarrow I=\int\frac{2sint.cost\left(1-sint\right)}{\sqrt{1-sin^2t}}dt=\int\frac{2sint.cost\left(1-sint\right)}{cost}dt=\int\left(2sint-2sin^2t\right)dt\)
\(=\int\left(2sint+cos2t-1\right)dt=-2cost+\frac{1}{2}sin2t-t+C\)
\(=-2\sqrt{1-sin^2t}+\frac{1}{2}sint\sqrt{1-sin^2t}-t+C\)
\(=-2\sqrt{1-x}+\frac{1}{2}\sqrt{x\left(1-x\right)}-arcsin\left(\sqrt{x}\right)+C\)