1,\(\int\frac{3+2x}{\sqrt{x}}dx\)
2,\(\int\frac{xdx}{\sqrt{5-4x^2}}\)
3,\(\int\frac{dx}{\left(2x-3\right)^2}\)
4,\(\int\frac{\left(1+\sqrt{x}\right)^{\frac{1}{4}}}{\sqrt{x}}dx\)
5,\(\int\frac{2+3x}{\sqrt{1+4x+3x^2}}dx\)
6,\(\int\frac{lnxdx}{x\left[1+\left(lnx\right)^2\right]}\)
7,\(\int\frac{e^{tan^{-1}x}}{1+x^2}dx\)
8,\(\int\frac{sin\sqrt{x}}{\sqrt{x}}dx\)
9,\(\int\frac{cos\left(lnx\right)}{x}dx\)
10,\(\int\frac{sin^{-1}x}{\sqrt{1-x^2}}dx\)
Moị người giúp em với ạ !!!!
\(I_1=\int\left(\frac{6}{2\sqrt{x}}+2\sqrt{x}\right)dx=6\sqrt{x}+\frac{4}{3}\sqrt{x^3}+C\)
\(I_2=\int\frac{xdx}{\sqrt{5-4x^2}}\)
Đặt \(\sqrt{5-4x^2}=t\Rightarrow5-4x^2=t^2\)
\(\Rightarrow-8xdx=2tdt\Rightarrow xdx=-\frac{1}{4}tdt\)
\(\Rightarrow I_2=\int\frac{-\frac{1}{4}t.dt}{t}=-\frac{1}{4}\int dt=-\frac{1}{4}t+C=-\frac{1}{4}\sqrt{5-4x^2}+C\)
\(I_3=\int\frac{dx}{\left(2x-3\right)^2}=\frac{1}{2}\int\frac{d\left(2x-3\right)}{\left(2x-3\right)^2}=-\frac{1}{2\left(2x-3\right)}+C\)
\(I_4=\int\frac{\left(1+\sqrt{x}\right)^{\frac{1}{4}}}{\sqrt{x}}dx\)
Đặt \(\left(1+\sqrt{x}\right)^{\frac{1}{4}}=t\Rightarrow\sqrt{x}=t^4-1\)
\(\Rightarrow x=t^8-2t^4+1\Rightarrow dx=\left(8t^7-8t^3\right)dt\)
\(\Rightarrow I_4=\int\frac{t\left(8t^7-8t^3\right)}{t^4-1}dt=\int\frac{8t^4\left(t^4-1\right)}{t^4-1}dt=\int8t^4dt\)
\(=\frac{8}{5}t^5+C=\frac{8}{5}\left(1+\sqrt{x}\right)^{\frac{5}{4}}+C\)
\(I_5=\frac{2+3x}{\sqrt{1+4x+3x^2}}dx\)
Đặt \(\sqrt{1+4x+3x^2}=t\Rightarrow1+4x+3x^2=t^2\)
\(\Leftrightarrow\left(4+6x\right)dx=2tdt\Rightarrow\left(2+3x\right)dx=t.dt\)
\(\Rightarrow I_5=\int\frac{t.dt}{t}=\int dt=t+C=\sqrt{1+4x+3x^2}+C\)
\(I_6=\int\frac{lnx.dx}{x\left(1+ln^2x\right)}\)
Đặt \(1+ln^2x=t\Rightarrow\frac{2lnx}{x}dx=dt\Rightarrow\frac{lnxdx}{x}=\frac{dt}{2}\)
\(\Rightarrow I_6=\int\frac{dt}{2t}=\frac{1}{2}ln\left|t\right|+C=\frac{1}{2}ln\left(1+ln^2x\right)+C\)
\(I_7=\int\frac{e^{tan^{-1}x}dx}{1+x^2}\)
Đặt \(tan^{-1}x=t\Rightarrow\frac{dx}{1+x^2}=dt\)
\(\Rightarrow I_7=\int e^tdt=e^t+C=e^{tan^{-1}x}+C\)
\(I_8=\int\frac{sin\sqrt{x}}{\sqrt{x}}dx\)
Đặt \(\sqrt{x}=t\Rightarrow\frac{dx}{2\sqrt{x}}=dt\Rightarrow\frac{dx}{\sqrt{x}}=2dt\)
\(\Rightarrow I_8=\int2sint.dt=-2cost+C=-2cos\left(\sqrt{x}\right)+C\)
\(I_9=\int\frac{cos\left(lnx\right)dx}{x}\)
Đặt \(lnx=t\Rightarrow\frac{dx}{x}=dt\)
\(\Rightarrow I_9=\int costdt=sint+C=sin\left(lnx\right)+C\)
\(I_{10}=\int\frac{sin^{-1}x}{\sqrt{1-x^2}}dx\)
Đặt \(sin^{-1}x=t\Rightarrow\frac{dx}{\sqrt{1-x^2}}=dt\)
\(\Rightarrow I_{10}=\int t.dt=\frac{1}{2}t^2+C=\frac{1}{2}\left(sin^{-1}x\right)^2+C\)
