Nguyên hàm bài tập phần 12

\( \int 3x \sqrt{7 - 3x^2} \, dx \)

Đặt \( t = \sqrt{7 - 3x^2} \Rightarrow 3x^2 = 7 - t^2 \)

\(\Rightarrow 6x \, dx = -2t \, dt \Rightarrow x \, dx = -\frac{1}{3} t \, dt \).

\( \int 3x \sqrt{7 - 3x^2} \, dx = \int -t^2 \, dt = -\frac{t^3}{3} + c \)

\( = -\frac{(7 - 3x^2) \sqrt{7 - 3x^2}}{3} + c \).

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\( \int \frac{e^{2x}}{\sqrt{1 + e^x}} \, dx \)

Đặt \( t = \sqrt{1 + e^x} \Rightarrow e^x = t^2 - 1 \)

\(\Rightarrow e^x dx = 2t dt \Rightarrow dx = \frac{2t}{t^2 - 1} \, dt \).

\( \int \frac{e^{2x}}{\sqrt{1 + e^x}} \, dx = \int \frac{\left( t^2 - 1 \right)^2}{t} \cdot \frac{2t}{(t^2 - 1)} \, dt \)

\( = \int 2(t^2 - 1) \, dt = 2 \frac{t^3}{3} - 2t + C \)

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\( \int \frac{\ln x \sqrt{1 + 3 \ln x}}{x} \, dx \quad (2004.B) \)

Đặt \( t = \sqrt{1 + 3 \ln x} \Rightarrow \ln x = \frac{1}{3}(t^2 - 1) \Rightarrow \frac{1}{x} dx = \frac{2}{3} t \, dt \).

\( \int \frac{\ln x \sqrt{1 + 3 \ln x}}{x} \, dx = \int \frac{1}{3} (t^2 - 1) t \cdot \frac{2}{3} t \, dt \)

\( = \frac{2}{9} \int (t^4 - t^2) \, dt \)

\( = \frac{2}{9} \left[ \frac{t^5}{5} - \frac{t^3}{3} \right] + C \).

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\( \int \frac{\sin 2x + \sin x}{\sqrt{1 + 3 \cos x}} \, dx \quad (2005.A) \)

\( \int \frac{\sin x (2 \cos x + 1)}{\sqrt{1 + 3 \cos x}} \, dx \)

Đặt \( t = \sqrt{1 + 3 \cos x} \Rightarrow \cos x = \frac{t^2 - 1}{3} \Rightarrow -\sin x \, dx = \frac{2}{3} t \, dt \).

\( \int \frac{\sin x (2 \cos x + 1)}{\sqrt{1 + 3 \cos x}} \, dx = \int -\frac{2}{3} t \left( \frac{2(t^2 - 1)}{3}  + 1 \right) \frac{1}{t} \, dt \)

\( = -\frac{2}{3} \int \frac{2(t^2 + 1)}{3} \, dt = -\frac{2}{9} \int (2t^2 + 1) \, dt \)

\( = -\frac{2}{9} \left[ \frac{2t^3}{3} + t \right] + C \)

\( = -\frac{4}{27} t^3 - \frac{2}{9} t + C \).

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*Vì \( \left( \frac{ax + b}{cx + d} \right)' = \frac{ad - bc}{(cx + d)^2} \).
Gặp: \( \int \frac{(ax + b)^n}{(cx + d)^{n+2}} \, dx \) thì đặt \( t = \frac{ax + b}{cx + d} \).

 

\( \int \frac{(2x - 1)^3}{(3x + 2)^5} \, dx \)

Đặt \( t = \frac{2x - 1}{3x + 2} \Rightarrow dt = \frac{7}{(3x + 2)^2} \, dx \).

\( \int \left[\frac{(2x - 1)^3}{(3x + 2)^5}\right]^3 \frac{1}{(3x + 2)^2} \, dx = \frac{1}{7} \int t^3 \, dt = \frac{t^4}{28} + C \).

Làm thêm:
\( \int \frac{(3x - 1)^5}{(4x + 3)^7} \, dx \)

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