Đáp án:

a) $\displaystyle\int\limits_0^1x\left(e^{\displaystyle{x}} + e^{\displaystyle{x^2}}\right)dx = \dfrac{e+1}{2}$

b) $\displaystyle\int\limits_0^{\ln8}\dfrac{x}{\sqrt{e^x+1}}=\ln\left(\dfrac{3+2\sqrt2}{2}\right)$

Giải thích các bước giải:

$\begin{array}{l}a)\quad I=\displaystyle\int\limits_0^1x\left(e^{\displaystyle{x}} + e^{\displaystyle{x^2}}\right)dx \\ = \displaystyle\int\limits_0^1xe^{\displaystyle{x^2}}dx+\displaystyle\int\limits_0^1xe^xdx\\ = I_1 + I_2\\ +) \quad I_1 = \displaystyle\int\limits_0^1xe^{\displaystyle{x^2}}dx\\ Đặt \,\,u = x^2\\ \to du = 2xdx\\ Đổi\,\,cận:\\ x\quad\vert \qquad 0\qquad \qquad 1\\ \hline u\quad \vert\qquad 0\qquad \qquad 1\\ \text{Ta được:}\\ I_1 = \dfrac12\displaystyle\int\limits_0^1e^udu\\ \to I_1 = \dfrac{e^u}{2}\Bigg|_0^1 = \dfrac{e^1 - e^0}{2} = \dfrac{e-1}{2}\\ Đặt\,\,\begin{cases}u = x\\dv =e^xdx \end{cases}\longrightarrow \begin{cases}du = dx\\v = e^x\end{cases}\\ \text{Ta được:}\\ I_2 = xe^x\Big|_0^1 - \displaystyle\int\limits_0^1e^xdx\\ \to I_2 = xe^x\Big|_0^1 - e^x\Big|_0^1\\ \to I_2 = 1.e^1 - 0.e^0 - (e^1 - e^0) = 1\\ Vậy\,\,I = I_1 + I_2 = \dfrac{e-1}{2} + 1 =\dfrac{e+1}{2}\\ b)\quad \displaystyle\int\limits_0^{\ln8}\dfrac{x}{\sqrt{e^x+1}}\\ Đặt\,\,u = e^x\\ \to du = e^xdx\\ Đổi\,\,cận:\\ x\quad\vert \qquad 0\qquad \qquad \ln8\\ \hline u\quad \vert\qquad 1\qquad \qquad 8\\ \text{Ta được:}\\ \displaystyle\int\limits_1^{8}\dfrac{du}{u\sqrt{u+1}}\\ Đặt\,\,t = u + 1\\ \to dt = du\\ Đổi\,\,cận:\\ u\quad\vert \qquad 1\qquad \qquad 8\\ \hline t\quad \vert\qquad 2\qquad \qquad 9\\ \text{Ta được:}\\ \displaystyle\int\limits_2^{9}\dfrac{dt}{(t-1)\sqrt t}\\ Đặt\,\,z = \sqrt t\\ \to dz = \dfrac{1}{2\sqrt t}dt\\ Đổi\,\,cận:\\ t\quad\vert \qquad 2\qquad \qquad 9\\ \hline z\quad \vert\qquad \sqrt2\qquad \qquad 3\\ \text{Ta được:}\\ -2\displaystyle\int\limits_\sqrt2^3\dfrac{dz}{1-z^2}\\ = -2\left[\left(\dfrac{\ln|1+z| - \ln|1-z|}{2}\right)\Bigg|_\sqrt2^3 \right]\\ =-2 \left[\dfrac{\ln4 - \ln2 - \ln(1+\sqrt2) + \ln(\sqrt2 -1)}{2} \right]\\ = \ln2 + \ln(1+\sqrt2) - \ln4 -\ln(\sqrt2-1)\\ = \ln\left(\dfrac{3+2\sqrt2}{2}\right) \end{array}$