|
giải đáp
|
nguyên hàm
|
|
|
$\int\limits (1+tanx)^2e^{2x}dx=\int\limits (1+tan^2x)e^{2x}dx+\int\limits 2e^{2x}tanxdx$ $=\int\limits (1+tan^2x)e^{2x}dx+\int\limits 2e^{2x}tanxdx$
$=e^{2x}tanx-\int\limits 2e^{2x}tanxdx+C+\int\limits 2e^{2x}tanxdx$
$=e^{2x}tanx + C$
|
|
|
|
giải đáp
|
nguyên hàm 12
|
|
|
$\int\limits \frac{1}{x^2-2x+3}dx=\int\limits \frac{1}{(x-1)^2+2}dx=\frac{1}{\sqrt{2}}arctan(\frac{x-1}{\sqrt{2}})+C$
|
|
|
giải đáp
|
nguyên hàm 12 (1)
|
|
|
$\int\limits \frac{3x+5}{2x^2+x+10}dx=\int\limits [\frac{3(4x+1)}{4(2x^2+x+10}+\frac{17}{4(2x^2+x+10)}]dx$ $=\int\limits [\frac{3(4x+1)}{4(2x^2+x+10)}+\frac{17}{8[(x+\frac{1}{4})^2+\frac{79}{16}]}]dx$
$=\frac{3}{4}ln(2x^2+x+10)+\frac{17}{2\sqrt{79}}artan\frac{4x+1}{\sqrt{79}}+C$
|
|
|
giải đáp
|
Nguyên hàm
|
|
|
$\int\limits \frac{1}{x^3+1}dx=\frac{1}{3}\int\limits( \frac{1}{x+1}-\frac{x-2}{x^2-x+1})dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2(x^2-x+1)}]dx$
$=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2}.\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}]dx$ $=\frac{1}{3}[ln|x+1|-\frac{1}{2}ln(x^2-x+1)+\sqrt{3}arctan(\frac{2x-1}{\sqrt{3}})]+C$
|
|
|
giải đáp
|
nguyên hàm
|
|
|
$\int\limits \frac{1}{cos2x(1+sin2x)}dx=\int\limits \frac{1}{(1-sin2x)(1+sin2x)^2}cos2xdx$.Đặt $t=1+sin2x$, có $cos2xdx=\frac{dt}{2}$. $\int\limits \frac{1}{(1-sin2x)(1+sin2x)^2}cos2xdx=\int\limits \frac{1}{(2-t)t^2}\frac{dt}{2}$.
|
|
|
giải đáp
|
Tính nguyên hàm: $I=\int \frac{dx}{sin^4x+cos^4x}$
|
|
|
$\int\limits \frac{1}{sin^4x+cos^4x}dx=\int\limits \frac{2}{2-sin^{2}2x}dx$ $=\int\limits \frac{2}{sin^{2}2x+2cos^{2}2x}dx$
$=\int\limits \frac{1}{tan^{2}2x+2}.\frac{2}{cos^{2}2x}dx$
$=\int\limits \frac{1}{tan^{2}2x+2}d(tan2x)$
$=\frac{1}{\sqrt{2}}arctan(\frac{tan2x}{\sqrt{2}})+C$
|
|
|
|
giải đáp
|
nguyên hàm
|
|
|
$\int\limits \frac{1}{1+sin^2x}dx=\int\limits \frac{1}{2sin^2x+cos^2x}dx$ $=\int\limits \frac{1}{2tan^2x+1}.\frac{1}{cos^2x}dx=$
$=\int\limits \frac{1}{2tan^2x+1}d(tanx)=$
$=\frac{\sqrt{2}}{2}arctan(\sqrt{2}tanx)+C$
|
|
|
giải đáp
|
Nguyên Hàm
|
|
|
$\int\limits \frac{1}{\sqrt{x-x^2}}dx=\int\limits \frac{1}{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}dx=arcsin(2x-1)+C$
|
|
|
giải đáp
|
Tính nguyên a
|
|
|
$\int\limits \frac{x}{e^x}dx=\int\limits xe^{-x}dx=-xe^{-x}-e^{-x}+C$
|
|
|
giải đáp
|
tính nguyên hàm
|
|
|
$\int\limits\frac{e^{\frac{1-2x}{x+1}}}{(x+1)^2}dx=-\frac{1}{3}\int\limits e^{\frac{1-2x}{x+1}}d(\frac{1-2x}{x+1})=-\frac{1}{3}e^{\frac{1-2x}{x+1}}+C$
|
|
|
giải đáp
|
nguyên hàm
|
|
|
$\int\limits x^2(x-9)^4dx=\int\limits [(x-9)+9]^2(x-9)^4dx$ $=\int\limits [(x-9)^6+18(x-9)^5+81(x-9)^4]dx$
$=\frac{(x-9)^7}{7}+3(x-9)^6+\frac{81(x-9)^5}{5}+C$.
|
|