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giải đáp
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nguyên hàm
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$\int\limits (1+tanx)^2e^{2x}dx=\int\limits (1+tan^2x)e^{2x}dx+\int\limits 2e^{2x}tanxdx$ $=\int\limits (1+tan^2x)e^{2x}dx+\int\limits 2e^{2x}tanxdx$
$=e^{2x}tanx-\int\limits 2e^{2x}tanxdx+C+\int\limits 2e^{2x}tanxdx$
$=e^{2x}tanx + C$
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giải đáp
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nguyên hàm 12
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$\int\limits \frac{1}{x^2-2x+3}dx=\int\limits \frac{1}{(x-1)^2+2}dx=\frac{1}{\sqrt{2}}arctan(\frac{x-1}{\sqrt{2}})+C$
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giải đáp
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nguyên hàm 12 (1)
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$\int\limits \frac{3x+5}{2x^2+x+10}dx=\int\limits [\frac{3(4x+1)}{4(2x^2+x+10}+\frac{17}{4(2x^2+x+10)}]dx$ $=\int\limits [\frac{3(4x+1)}{4(2x^2+x+10)}+\frac{17}{8[(x+\frac{1}{4})^2+\frac{79}{16}]}]dx$
$=\frac{3}{4}ln(2x^2+x+10)+\frac{17}{2\sqrt{79}}artan\frac{4x+1}{\sqrt{79}}+C$
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sửa đổi
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Nguyên hàm
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$\int\limits \frac{1}{x^3+1}dx=\frac{1}{3}\int\limits( \frac{1}{x+1}-\frac{x-2}{x^2-x+1})dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2(x^2-x+1)}]dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2}.\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}]dx$ $\frac{1}{3}[ln|x+1|-\frac{1}{2}ln(x^2-x+1)+\sqrt{3}arctan(\frac{2x-1}{\sqrt{3}})]+C$
$\int\limits \frac{1}{x^3+1}dx=\frac{1}{3}\int\limits( \frac{1}{x+1}-\frac{x-2}{x^2-x+1})dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2(x^2-x+1)}]dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2}.\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}]dx$ $=\frac{1}{3}[ln|x+1|-\frac{1}{2}ln(x^2-x+1)+\sqrt{3}arctan(\frac{2x-1}{\sqrt{3}})]+C$
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giải đáp
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Nguyên hàm
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$\int\limits \frac{1}{x^3+1}dx=\frac{1}{3}\int\limits( \frac{1}{x+1}-\frac{x-2}{x^2-x+1})dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2(x^2-x+1)}]dx$
$=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2}.\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}]dx$ $=\frac{1}{3}[ln|x+1|-\frac{1}{2}ln(x^2-x+1)+\sqrt{3}arctan(\frac{2x-1}{\sqrt{3}})]+C$
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giải đáp
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nguyên hàm
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$\int\limits \frac{1}{cos2x(1+sin2x)}dx=\int\limits \frac{1}{(1-sin2x)(1+sin2x)^2}cos2xdx$.Đặt $t=1+sin2x$, có $cos2xdx=\frac{dt}{2}$. $\int\limits \frac{1}{(1-sin2x)(1+sin2x)^2}cos2xdx=\int\limits \frac{1}{(2-t)t^2}\frac{dt}{2}$.
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giải đáp
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Tính nguyên hàm: $I=\int \frac{dx}{sin^4x+cos^4x}$
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$\int\limits \frac{1}{sin^4x+cos^4x}dx=\int\limits \frac{2}{2-sin^{2}2x}dx$ $=\int\limits \frac{2}{sin^{2}2x+2cos^{2}2x}dx$
$=\int\limits \frac{1}{tan^{2}2x+2}.\frac{2}{cos^{2}2x}dx$
$=\int\limits \frac{1}{tan^{2}2x+2}d(tan2x)$
$=\frac{1}{\sqrt{2}}arctan(\frac{tan2x}{\sqrt{2}})+C$
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