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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 \(t=\sqrt{1+\sqrt[3]{x^2}}\Rightarrow t^2-1=\sqrt[3]{x^2}\Leftrightarrow x^2=\left(t^2-1\right)^3\)
\(d\left(x^2\right)=d\left[\left(t^2-1\right)^3\right]\Leftrightarrow2x\cdot dx=6t\left(t^2-1\right)^2\cdot dt\)
\(I_3=\int\frac{3t\left(t^2-1\right)^2}{t}dt=3\int\left(t^4-2t^2+1\right)dt=...\)
Câu 1)
Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).
Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)
Khi đó:
\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)
Câu 3)
\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)
\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\): Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)
\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)
\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)
Bài 2)
\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)
Khi đó:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
a) Đặt \(\sqrt{2x-5}=t\) khi đó \(x=\frac{t^2+5}{2}\) , \(dx=tdt\)
Do vậy \(I_1=\int\frac{\frac{1}{4}\left(t^2+5\right)^2+3}{t^3}dt=\frac{1}{4}\int\frac{\left(t^4+10t^2+37\right)t}{t^3}dt\)
\(=\frac{1}{4}\int\left(t^2+10+\frac{37}{t^2}\right)dt=\frac{1}{4}\left(\frac{t^3}{3}+10t-\frac{37}{t}\right)+C\)
Trở về biến x, thu được :
\(I_1=\frac{1}{12}\sqrt{\left(2x-5\right)^3}+\frac{5}{2}\sqrt{2x-5}-\frac{37}{4\sqrt{2x-5}}+C\)
b) \(I_2=\frac{1}{3}\int\frac{d\left(\ln\left(3x-1\right)\right)}{\ln\left(3x-1\right)}=\frac{1}{3}\ln\left|\ln\left(3x-1\right)\right|+C\)
c) \(I_3=\int\frac{1+\frac{1}{x^2}}{\sqrt{x^2-7+\frac{1}{x^2}}}dx=\int\frac{d\left(x-\frac{1}{x}\right)}{\sqrt{\left(x-\frac{1}{2}\right)^2-5}}\)
Đặt \(x-\frac{1}{x}=t\)
\(\Rightarrow\) \(I_3=\int\frac{dt}{\sqrt{t^2-5}}=\ln\left|t+\sqrt{t^2-5}\right|+C\)
\(=\ln\left|x-\frac{1}{x}+\sqrt{x^2-7+\frac{1}{x^2}}\right|+C\)
14.
\(log_aa^2b^4=log_aa^2+log_ab^4=2+4log_ab=2+4p\)
15.
\(\frac{1}{2}log_ab+\frac{1}{2}log_ba=1\)
\(\Leftrightarrow log_ab+\frac{1}{log_ab}=2\)
\(\Leftrightarrow log_a^2b-2log_ab+1=0\)
\(\Leftrightarrow\left(log_ab-1\right)^2=0\)
\(\Rightarrow log_ab=1\Rightarrow a=b\)
16.
\(2^a=3\Rightarrow log_32^a=1\Rightarrow log_32=\frac{1}{a}\)
\(log_3\sqrt[3]{16}=log_32^{\frac{4}{3}}=\frac{4}{3}log_32=\frac{4}{3a}\)
11.
\(\Leftrightarrow1>\left(2+\sqrt{3}\right)^x\left(2+\sqrt{3}\right)^{x+2}\)
\(\Leftrightarrow\left(2+\sqrt{3}\right)^{2x+2}< 1\)
\(\Leftrightarrow2x+2< 0\Rightarrow x< -1\)
\(\Rightarrow\) có \(-2+2020+1=2019\) nghiệm
12.
\(\Leftrightarrow\left\{{}\begin{matrix}x-2>0\\0< log_3\left(x-2\right)< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\1< x-2< 3\end{matrix}\right.\)
\(\Rightarrow3< x< 5\Rightarrow b-a=2\)
13.
\(4^x=t>0\Rightarrow t^2-5t+4\ge0\)
\(\Rightarrow\left[{}\begin{matrix}t\le1\\t\ge4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}4^x\le1\\4^x\ge4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le0\\x\ge1\end{matrix}\right.\)
Câu 1:
Đặt \(\sqrt{lnx+1}=t\Rightarrow lnx=t^2-1\Rightarrow\frac{dx}{x}=2tdt\)
\(\Rightarrow I=\int3t.2t.dt=6\int t^2dt=2t^3+C\)
\(=2\sqrt{\left(lnx+1\right)^3}+C=2\left(lnx+1\right)\sqrt{lnx+1}+C\)
\(=ln\left(x.e\right)^2\sqrt{ln\left(x.e\right)+0}\Rightarrow a=2;b=0\)
Câu 2:
\(\int\limits^b_ax^{-\frac{1}{2}}dx=2x^{\frac{1}{2}}|^b_a=2\left(\sqrt{b}-\sqrt{a}\right)=2\Rightarrow\sqrt{b}-\sqrt{a}=1\)
Ta có hệ: \(\left\{{}\begin{matrix}\sqrt{b}-\sqrt{a}=1\\a^2+b^2=17\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=4\\a=1\end{matrix}\right.\) (lưu ý loại cặp nghiệm âm do \(\frac{1}{\sqrt{x}}\) chỉ xác định trên miền (a;b) dương)
Câu 4:
\(\int\frac{3x+a}{x^2+4}dx=\frac{3}{2}\int\frac{2x}{x^2+4}dx+a\int\frac{1}{x^2+4}dx\)
\(=\frac{3}{2}ln\left(x^2+4\right)+\frac{a}{2}arctan\left(\frac{x}{2}\right)+C\)
\(\Rightarrow a=2\)
\(\Rightarrow I=\int\limits^{\frac{e}{4}}_1ln\left(x\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{1}{x}dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I=x.lnx|^{\frac{e}{4}}_1-\int\limits^{\frac{e}{4}}_1dx=\frac{e}{4}.ln\left(\frac{e}{4}\right)-\frac{e}{4}+1=-\frac{ln\left(2^e\right)}{2}+1\)
Câu 5:
\(f'\left(x\right)=\int f''\left(x\right)dx=-\frac{1}{4}\int x^{-\frac{3}{2}}dx=\frac{1}{2\sqrt{x}}+C\)
\(f'\left(2\right)=\frac{1}{2\sqrt{2}}+C=2+\frac{1}{2\sqrt{2}}\Rightarrow C=2\)
\(\Rightarrow f'\left(x\right)=\frac{1}{2\sqrt{x}}+2\)
\(\Rightarrow f\left(x\right)=\int f'\left(x\right)dx=\int\left(\frac{1}{2\sqrt{x}}+2\right)dx=\sqrt{x}+2x+C_1\)
\(f\left(4\right)=\sqrt{4}+2.4+C_1=10\Rightarrow C_1=0\)
\(\Rightarrow f\left(x\right)=2x+\sqrt{x}\)
\(\Rightarrow F\left(x\right)=\int f\left(x\right)dx=\int\left(2x+\sqrt{x}\right)dx=x^2+\frac{2}{3}\sqrt{x^3}+C_2\)
\(F\left(1\right)=1+\frac{2}{3}+C_2=1+\frac{2}{3}\Rightarrow C_2=0\)
\(\Rightarrow F\left(x\right)=x^2+\frac{2}{3}\sqrt{x^3}\Rightarrow\int\limits^1_0\left(x^2+\frac{2}{3}\sqrt{x^3}\right)dx=\frac{3}{5}\)
\(\left(\frac{1}{4}x-1\right)-\left(\frac{2}{3}x-1\right)+\left(\frac{4}{5}x-1\right)=\frac{2}{3}\)
\(\frac{1}{4}x-1-\frac{2}{3}x+1+\frac{4}{5}x-1\)\(=\frac{2}{3}\)
\(\left(\frac{1}{4}x-\frac{2}{3}x+\frac{4}{5}x\right)+1-1-1\)\(=\frac{2}{3}\)
\(\frac{23}{60}x-1\)\(=\frac{2}{3}\)
\(\frac{23}{60}x=\frac{2}{3}+1\)
\(\frac{23}{60}x=\frac{5}{3}\)
\(x=\frac{5}{3}:\frac{23}{60}=\frac{100}{23}\)
Vậy x=\(\frac{100}{23}\)
Giải:
Ta có: \(\frac{x-2}{5}=\frac{2x-3}{4}\)
\(\Rightarrow\left(x-2\right).4=5.\left(2x-3\right)\)
\(\Rightarrow4x-8=10x-15\)
\(\Rightarrow4x-10x=8-15\)
\(\Rightarrow-6x=-7\)
\(\Rightarrow x=\frac{7}{6}\)
Vậy \(x=\frac{7}{6}\)
Giải :
Ta có : \(\frac{x-2}{5}=\frac{2x-3}{4}\)
\(\Rightarrow\left(x-2\right),4=5,\left(2x-3\right)\)
\(\Rightarrow4x-8=10x-15\)
\(\Rightarrow4x-10x=8-15\)
\(\Rightarrow-6x=-7\)
\(\Rightarrow x=\frac{7}{6}\)
Vậy \(x\) là \(\frac{7}{6}\)