Bài tích phân

Question

1, $\int\limits_{0}^{\frac{\Pi }{4}}ln(1+tanx)dx$

2, $\int\limits_{0}^{ln2}\frac{e^{2x}}{\sqrt{e^x+1}}dx$

3, $\int\limits_{0}^{3}\frac{x^2+1}{\sqrt{x+1}}dx$

score 3 · Answer 1

Bình chọn tăng 3 Bình chọn giảm

Câu 3

$I = \int_0^3 \dfrac{(x+1)^2}{\sqrt{x+1}}dx -2\int \dfrac{x}{\sqrt{x+1}}dx$

$=\int_0^3 (x+1)^{\frac{3}{2}}dx -2\int_0^3 \dfrac{x}{\sqrt{x+1}}dx$ bạn tính giống hệt bà 2 mình chữa bên trên nhé

score 2 · Answer 2

Bình chọn tăng 2 Bình chọn giảm

Câu 2

$I=\int_0^{\ln 2} \dfrac{e^x .e^x}{\sqrt{e^x+1}}dx = \int_0^{\ln 2} \dfrac{e^x d(e^x)}{\sqrt{e^x+1}} = \int_0^2 \dfrac{t}{\sqrt{t+1}}dt$

$=\int_0^2 \dfrac{t+1}{\sqrt{t+1}}dt - \int_0^2 \dfrac{1}{\sqrt{t+1}}dt = \int_0^2 (t+1)^{\frac{1}{2}}d(t+1) -\int_0^2 \dfrac{d(t+1)}{(t+1)^{\frac{1}{2}}}$

$=\dfrac{2}{3}. \sqrt{(t+1)^3} \bigg |_0^2 - \int_0^2 {(t+1)^{-\frac{1}{2}}} d(t+1) = \dfrac{2}{3}. \sqrt{(t+1)^3} \bigg |_0^2 +\dfrac{1}{2\sqrt{t+1}}\bigg |_0^2$

score 2 · Answer 3

Bình chọn tăng 2 Bình chọn giảm

Câu a. đặt $x=\dfrac{\pi}{4} -t\Rightarrow dx = -dt$

$I=-\int_{\frac{\pi}{4}}^0 \ln (1+\tan (\dfrac{\pi}{4}-t)dt = \int_0^\frac{\pi}{4} \ln (1+\dfrac{1-\tan t}{1+\tan t})dt$

$= \int_0^\frac{\pi}{4} \ln (\dfrac{2}{1+\tan t})dt = \int_0^\frac{\pi}{4} [\ln 2 - \ln (1+\tan t)]dt = \int_0^\frac{\pi}{4} \ln 2 dt- \int_0^\frac{\pi}{4} ln (1+\tan t)dt =\ln 2 t \bigg |_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} ln (1+\tan x)dx = \dfrac{\pi}{4}\ln 2 -I$

$\Rightarrow 2I = \dfrac{\pi}{4}\ln 2 \Rightarrow I= \dfrac{\pi}{8}\ln 2$