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giải đáp
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Tích phân thi Đại học(5).
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Ta có: $\int\limits_0^{\frac{\pi}{4}}x(1+\sin2x)dx$ $=\int\limits_0^{\frac{\pi}{4}}xdx+\int\limits_0^{\frac{\pi}{4}}x\sin2xdx$ $=\dfrac{x^2}{2}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.+\int\limits_0^{\frac{\pi}{4}}xd(\sin^2x)$ $=\dfrac{\pi^2}{32}+x\sin^2x\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.-\int\limits_0^{\frac{\pi}{4}}\sin^2xdx$ $=\dfrac{\pi^2}{32}+\dfrac{\pi}{8}-\dfrac{1}{2}\int\limits_0^{\frac{\pi}{4}}(1-\cos2x)dx$ $=\dfrac{\pi^2}{32}+\dfrac{\pi}{8}-\dfrac{2x-\sin2x}{4}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.$ $=\dfrac{\pi^2}{32}+\dfrac{1}{4}$ |
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giải đáp
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Tích phân thi Đại học(6).
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Ta có: $\int\limits_0^{\frac{\pi}{4}}\dfrac{x\sin x+(x+1)\cos x}{x\sin x+c\cos x}dx$ $=\int\limits_0^{\frac{\pi}{4}}\left(1+\dfrac{x\cos x}{x\sin x+\cos x}\right)dx$ $=\int\limits_0^{\frac{\pi}{4}}dx+\int\limits_0^{\frac{\pi}{4}}\dfrac{d(x\sin x+\cos x)}{x\sin x+\cos x}$ $=\left(x+\ln(x\sin x+\cos x)\right)\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.$ $=\dfrac{\pi}{4}+\ln\dfrac{\pi+4}{4\sqrt2}$ |
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giải đáp
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Tích phân thi Đại học(10).
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Ta có: $\int\limits_0^1\dfrac{x^2+e^x+2x^2e^x}{1+2e^x}dx$ $=\int\limits_0^1\left(x^2+\dfrac{e^x}{1+2e^x}\right)dx$ $=\left(\dfrac{x^3}{3}+\dfrac{1}{2}\ln(1+2e^x)\right)\left|\begin{array}{l}1\\0\end{array}\right.$ $=\dfrac{1}{3}+\dfrac{1}{2}\ln\dfrac{1+2e}{3}$ |
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giải đáp
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Tích phân thi Đại học(9).
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Ta có: $\int\limits_1^2\dfrac{2x+1}{x(x+1)}dx$ $=\int\limits_1^2\left(\dfrac{1}{x}+\dfrac{1}{x+1}\right)dx$ $=\ln[x(x+1)]\left|\begin{array}{l}2\\1\end{array}\right.$ $=\ln3$ |
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sửa đổi
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Tích phân thi Đại học(20).
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Ta có: $\int\limits_1^2\dfrac{\ln x}{x^3}dx$$=-\dfrac{1}{2}\int\limits_1^2\ln xd\left(\dfrac{1}{x^2}\right)$$=-\dfrac{\ln x}{2x^2}\left|\begin{array}{l}2\\1\end{array}\right.-\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^2}d(\ln x)$$=-\dfrac{\ln2}{8}-\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^3}dx$$=-\dfrac{\ln2}{8}+\dfrac{1}{4x^2}\left|\begin{array}{l}2\\1\end{array}\right.$$=-\dfrac{\ln2}{8}-\dfrac{3}{16}$
Ta có: $\int\limits_1^2\dfrac{\ln x}{x^3}dx$$=-\dfrac{1}{2}\int\limits_1^2\ln xd\left(\dfrac{1}{x^2}\right)$$=-\dfrac{\ln x}{2x^2}\left|\begin{array}{l}2\\1\end{array}\right.+\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^2}d(\ln x)$$=-\dfrac{\ln2}{8}+\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^3}dx$$=-\dfrac{\ln2}{8}-\dfrac{1}{4x^2}\left|\begin{array}{l}2\\1\end{array}\right.$$=-\dfrac{\ln2}{8}+\dfrac{3}{16}$
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giải đáp
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Tích phân thi Đại học(13).
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Ta có: $\int\limits_0^1\dfrac{2x-1}{x+1}dx$ $=\int\limits_0^1\left(2-\dfrac{3}{x+1}\right)dx$ $=(2x-3\ln(x+1))\left|\begin{array}{l}1\\0\end{array}\right.$ $=2-3\ln2$ |
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giải đáp
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Tích phân thi Đại học(17).
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Đặt: $t=e^x \Rightarrow dt=e^xdx$ Ta có: $\int\limits_1^3\dfrac{dx}{e^x-1}$ $=\int\limits_1^3\dfrac{e^xdx}{e^{2x}-e^x}$ $=\int\limits_e^{e^3}\dfrac{dt}{t^2-t}$ $=\int\limits_e^{e^3}\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt$ $=\ln\left|\dfrac{t-1}{t}\right|\left|\begin{array}{l}e^3\\e\end{array}\right.$ $=\ln\dfrac{e^2+e+1}{e^2}$ |
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giải đáp
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Tích phân thi Đại học(20).
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Ta có: $\int\limits_1^2\dfrac{\ln x}{x^3}dx$ $=-\dfrac{1}{2}\int\limits_1^2\ln xd\left(\dfrac{1}{x^2}\right)$ $=-\dfrac{\ln x}{2x^2}\left|\begin{array}{l}2\\1\end{array}\right.+\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^2}d(\ln x)$ $=-\dfrac{\ln2}{8}+\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^3}dx$ $=-\dfrac{\ln2}{8}-\dfrac{1}{4x^2}\left|\begin{array}{l}2\\1\end{array}\right.$ $=-\dfrac{\ln2}{8}+\dfrac{3}{16}$ |
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giải đáp
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Tích phân thi Đại học(22).
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Đặt: $t=\sin^2x \Rightarrow dt=\sin2xdx$ Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x}{\sqrt{\cos^2x+4\sin^2x}}dx$ $=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x}{\sqrt{1+3\sin^2x}}dx$ $=\int\limits_0^1\dfrac{dt}{\sqrt{1+3t}}$ $=\dfrac{1}{3}\int\limits_0^1(1+3t)^{\frac{-1}{2}}d(1+3t)$ $=\dfrac{2}{3}(1+3t)^{\frac{1}{2}}\left|\begin{array}{l}1\\0\end{array}\right.$ $=\dfrac{2}{3}$ |
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giải đáp
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Tích phân thi Đại học(23).
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Đặt: $t=e^x \Rightarrow dt=e^xdx$ Ta có: $\int\limits_{\ln 3}^{\ln 5}\dfrac{dx}{e^x+2e^{-x}+3}$ $=\int\limits_{\ln 3}^{\ln5}\dfrac{e^xdx}{e^{2x}-3e^x+2}$ $=\int\limits_3^5\dfrac{dt}{t^2-3t+2}$ $=\int\limits_3^5\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)dt$ $=\ln\left|\dfrac{t-2}{t-1}\right|\left|\begin{array}{l}5\\3\end{array}\right.$ $=\ln\dfrac{3}{2}$ |
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giải đáp
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Tích phân thi Đại học(25).
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Đặt: $t=\sqrt{1+3\cos x} \Rightarrow t^2=1+3\cos x\Rightarrow 2tdt=-3\sin xdx$ Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx$ $=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x(2\cos x+1)}{\sqrt{1+3\cos x}}dx$ $=-\dfrac{2}{3}\int\limits_2^1\dfrac{2\dfrac{t^2-1}{3}+1}{t}tdt$ $=\dfrac{2}{9}\int\limits_1^2(2t^2+1)dt$ $=\dfrac{2}{9}\left(\dfrac{2t^3}{3}+t\right)\left|\begin{array}{l}2\\1\end{array}\right.$ $=\dfrac{34}{27}$ |
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giải đáp
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Tích phân thi Đại học(26).
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Đặt: $t=1+\cos x \Rightarrow dt=-\sin xdx$ Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x\cos x}{1+\cos x}dx$ $=2\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x\cos^2x}{1+\cos x}dx$ $=-2\int\limits_2^1\dfrac{(t-1)^2}{t}dt$ $=2\int\limits_1^2\dfrac{(t-1)^2}{t}dt$ $=2\int\limits_1^2\left(t-2+\dfrac{1}{t}\right)dt$ $=2\left(\dfrac{t^2}{2}-2t+\ln t\right)\left|\begin{array}{l}2\\1\end{array}\right.$ $=2\ln2-1$ |
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