# Câu 14 trang 213 SGK Giải tích 12 Nâng cao

##### Hướng dẫn giải

a) Đặt $$x = \tan t \Rightarrow dx = {1 \over {{{\cos }^2}t}}dt$$

$$\int\limits_0^1 {{{dx} \over {{x^2} + 1}}} = \int\limits_0^{{\pi \over 4}} {{{dt} \over {{{\cos }^2}t({{\tan }^2}t + 1)}}} = \int\limits_0^{{\pi \over 4}} {dt} = {\pi \over 4}$$

b) Ta có:

$$I = \int\limits_0^1 {{{dx} \over {{x^2} + x + 1}}} = \int\limits_0^1 {{{dx} \over {{{(x + {1 \over 2})}^2} + {{({{\sqrt 3 } \over 2})}^2}}}}$$

Đặt $$x + {1 \over 2} = {{\sqrt 3 } \over 2}\tan t \Rightarrow dx = {{\sqrt 3 } \over 2}(1 + {\tan ^2}t)dt$$

$$I = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{{{\sqrt 3 } \over 2}dt} \over {{3 \over 4}}}} = {4 \over 3}.{{\sqrt 3 } \over 2}.{\pi \over 6} = {{\sqrt 3 \pi } \over 9}$$

c) Đặt

$$\left\{ \matrix{ u = {x^2} \hfill \cr dv = {e^x}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = 2xdx \hfill \cr v = {e^x} \hfill \cr} \right.$$

Do đó: $$\int\limits_0^1 {{x^2}{e^x}dx} = {x^2}{e^x}|_0^1 - 2\int\limits_0^1 {x{e^x}dx = e - 2\int\limits_0^1 {x{e^x}dx\,\,\,\,\,\,\,(*)} }$$

Đặt

$$\left\{ \matrix{ u = x \hfill \cr dv = {e^x}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = {e^x} \hfill \cr} \right.$$

Suy ra:

$$\int\limits_0^1 {x{e^x}dx = x{e^x}|_0^1} - \int\limits_0^1 {{e^x}dx} = e - {e^x}|_0^1 = 1$$

Từ (*) suy ra:  $$\int\limits_0^1 {{x^2}{e^x}dx} = e - 2$$