Đặt: \( \begin{cases} u = x \\ dv = \frac{e^x}{(1 + e^x)^2} \, dx \end{cases} \quad \Rightarrow \quad \begin{cases} du = dx \\ v = -\frac{1}{1 + e^x} \end{cases} \)
\( I = -\frac{x}{1 + e^x} + \int \frac{1}{1 + e^x} \, dx \)
\( = -\frac{x}{1 + e^x} + \int \frac{1 + e^x - e^x}{1 + e^x} \, dx \)
\( = -\frac{x}{1 + e^x} + \int \left( 1 - \frac{e^x}{1 + e^x} \right) \, dx \)
\( = -\frac{x}{1 + e^x} + x - \ln(1 + e^x) + C \)
page95
Đặt \( t = \sqrt{x}\Rightarrow x = t^2 \Rightarrow dx = 2t \, dt \).
\( I = 2 \int t e^t \, dt \)
\( = 2 \left[ t e^t - e^t \right] + C \)
\( = 2 \left[ \sqrt{x} e^{\sqrt{x}} - e^{\sqrt{x}} \right] + C \).
page96
Có thể đặt \( u = e^{2x} \) hoặc \( u = \sin{3x} \).
Đặt: \( \begin{cases} u = e^{2x} \\ dv = \sin{3x} \, dx \end{cases} \quad \Rightarrow \quad \begin{cases} du = 2e^{2x} \, dx \\ v = -\frac{1}{3} \cos{3x} \end{cases} \)
\( I = -\frac{1}{3} e^{2x} \cos{3x} + \frac{2}{3} \int e^{2x} \cos{3x} \, dx \)
Đặt: \( \begin{cases} u = e^{2x} \\ dv = \cos{3x} \, dx \end{cases} \quad \Rightarrow \quad \begin{cases} du = 2e^{2x} \, dx \\ v = \frac{1}{3} \sin{3x} \end{cases} \)
\( I = -\frac{1}{3} e^{2x} \cos{3x} + \frac{2}{3}\left[\frac{1}{3} e^{2x} \sin{3x} - \frac{2}{3}I\right] \)
\( I + \frac{4}{9} I = -\frac{1}{3} e^{2x} \cos{3x} + \frac{2}{9} e^{2x} \sin{3x} \)
\( \Rightarrow I = \frac{9}{13} \left[ \frac{2}{9} e^{2x} \sin{3x} - \frac{1}{3} e^{2x} \cos{3x} \right] + C \)
page97
\( I = \int 2 \cos{x} \cdot \sin{x} e^{\sin{x}} \, dx = 2 \int \left( u' f(u) \, dx \right) \quad \text{với } u = \sin{x} \quad \text{hoặc} \int \cos x f(\sin x) \, dx . \)
Đặt: \( t = \sin{x}, \quad \text{suy ra } dt = \cos{x} \, dx. \)
\( I = 2 \int t e^t \, dt. \)
Đặt: \( \begin{cases} u = t \\ dv = e^t \, dt \end{cases} \quad \Rightarrow \quad \begin{cases} du = dt \\ v = e^t \end{cases} \)
\( I = 2 \left[ t e^t - \int e^t \, dt \right] = 2 \left[ t e^t - e^t \right] + C \)
\( = 2 \left[ \sin{x} e^{\sin{x}} - e^{\sin{x}} \right] + C. \)
page98
Gợi ý: \( \left( \sin^2{x} \right)' = 2 \sin{x} \cos{x}. \)
\( I = \int \sin{x} \cos{x} \left( 1 - \sin^2{x} \right) e^{\sin^2{x}} \, dx. \)
\( = \frac{1}{2} \int u' f(u) \, du \quad \text{với } u = \sin^2{x}. \)
Đặt \( t = \sin^2{x} \Rightarrow dt = 2 \sin{x} \cos{x} \, dx \).
\( I = \frac{1}{2} \int (1 - t) e^t \, dt. \)
Đặt: \( \begin{cases} u = 1 - t \\ dv = e^t \, dt \end{cases} \quad \Rightarrow \quad \begin{cases} du = -dt \\ v = e^t \end{cases} \)
\( I = \frac{1}{2} \left[ (1 - t) e^t + \int e^t \, dt \right] \)
\( = \frac{1}{2} \left[ (1 - t) e^t + e^t \right] + C \)
\( = \frac{1}{2} \left[ (1 - \sin^2{x}) e^{\sin^2{x}} + e^{\sin^2{x}} \right] + C \)
page99