Đáp án:

$\begin{array}{l}1)\quad \dfrac{9}{20}\\ 2)\quad \dfrac{251}{12}\\ 3)\quad \dfrac{1}{168}\\ 4)\quad \dfrac43\\ 5)\quad \ln2\\ 6)\quad \dfrac{65}{4}\\ 7)\quad \dfrac{15}{16}\\ 8)\quad \dfrac{8\sqrt2 - 7}{15}\\ 9)\quad \dfrac18\\ 10)\quad \dfrac{4\sqrt2 - 2}{3}\\ 11)\quad \dfrac{\pi}{8} - \dfrac14\\ 12)\quad \dfrac{1}{2010}\\ 13)\quad \dfrac14\ln\dfrac53\\ 14)\quad \dfrac13\\ 15)\quad 3 - \sqrt3\\ 16)\quad 1\end{array}$

Giải thích các bước giải:

$\begin{array}{l}1)\quad I = \displaystyle\int\limits_0^1x^3(x+1)\\
\to I = \displaystyle\int\limits_0^1(x^4 + x^3)dx\\
\to I = \displaystyle\int\limits_0^1x^4dx + \displaystyle\int\limits_0^1x^3dx\\
\to I = \dfrac{x^5}{5}\Bigg|_0^1 + \dfrac{x^4}{4}\Bigg|_0^1\\
\to I = \dfrac15 + \dfrac14\\
\to I = \dfrac{9}{20}\\
2)\quad I = \displaystyle\int\limits_2^4\left(x + \dfrac1x\right)^2dx\\
\to I = \displaystyle\int\limits_2^4\left(x^2 + \dfrac{1}{x^2} + 2\right)dx\\
\to I = \displaystyle\int\limits_2^4x^2dx + \displaystyle\int\limits_2^4\dfrac{1}{x^2}dx + 2\displaystyle\int\limits_2^4dx\\
\to I = \dfrac{x^3}{3}\Bigg|_2^4 - \dfrac{1}{x}\Bigg|_2^4 + x\Bigg|_2^4\\
\to I = \left(\dfrac{64}{3} - \dfrac{8}{3}\right) - \left(\dfrac{1}{4} - \dfrac{1}{2}\right) + (4 - 2)\\
\to I = \dfrac{251}{12}\\
3)\quad I = \displaystyle\int\limits_0^1x^5(1-x^3)^6dx\\
Đặt\,\,u = 1-x^3\\
\to du = -3x^2dx\\
\text{Đổi cận:}\\
x\quad \Big|\quad 0 \qquad 1\\
\overline{u\quad \Big|\quad 1\qquad 0}\\
\text{Ta được:}\\
\quad I = -\dfrac13\displaystyle\int\limits^0_1u^6(1-u)du\\
\to I = \dfrac13\displaystyle\int\limits^0_1(u^7 - u^6)du\\
\to I = \dfrac13\displaystyle\int\limits_0^1u^6du - \dfrac13\displaystyle\int\limits_0^1u^7du\\
\to I = \dfrac{u^7}{21}\Bigg|_0^1 - \dfrac{u^8}{24}\Bigg|_0^1\\
\to I = \dfrac{1}{21} - \dfrac{1}{24}\\
\to I = \dfrac{1}{168}\\
4)\quad I = \displaystyle\int\limits_0^\sqrt3\dfrac{x^3dx}{\sqrt{x^2+1}}\\
Đặt\,\,u = x^2+1\\
\to du = 2xdx\\
\text{Đổi cận:}\\
x\quad \Big|\quad 0 \qquad \sqrt3\\
\overline{u \quad \Big|\quad 1 \qquad 4\quad}\\
\text{Ta được:}\\
\quad I = \dfrac12\displaystyle\int\limits_1^4\dfrac{u -1}{\sqrt u}du\\
\to I = \dfrac12\displaystyle\int\limits_1^4\sqrt udu - \dfrac12\displaystyle\int\limits_1^4\dfrac{du}{\sqrt u}\\
\to I = \dfrac{\sqrt{u^3}}{3}\Bigg|_1^4 - \sqrt u\Bigg|_1^4\\
\to I = \dfrac73 - 1\\
\to I = \dfrac43\\
5)\quad I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\dfrac{\sin xdx}{1+\cos x}\\
Đặt\,\,u = 1 + \cos x\\
\to du = -\sin xdx\\
\text{Đổi cận:}\\
x\quad \Big|\quad 0 \qquad \dfrac{\pi}{2}\\
\overline{u\quad \Big| \quad 2\qquad \,\,1}\\
\text{Ta được:}\\
\quad I = -\displaystyle\int\limits_2^1\dfrac1udu\\
\to I = \displaystyle\int\limits^2_1\dfrac1udu\\
\to I = \ln|u|\Bigg|_1^2\\
\to I  = \ln2\\
6)\quad I = \displaystyle\int\limits_1^{\tfrac{22}{3}}\sqrt[3]{3x+5}dx\\
Đặt\,\,u = 3x+5\\
\to du= 3dx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 1\qquad \dfrac{22}{3}\\
\overline{u\quad \Big|\quad 8\qquad \,\,27}\\
\text{Ta được:}\\
\quad I = \dfrac13\displaystyle\int\limits_8^{27}\sqrt[3]{u}du\\
\to I = \dfrac{\sqrt[3]{u^4}}{4}\Bigg|_8^{27}\\
\to I = \dfrac{\sqrt[3]{27^4} - \sqrt[3]{8^4}}{4}\\
\to I = \dfrac{65}{4}\\
7)\quad I = \displaystyle\int\limits_0^1x^3(1+x^4)^3dx\\
Đặt\,\,u = 1 + x^4\\
\to du = 4x^3dx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 0 \qquad 1\\
\overline{u\quad \Big|\quad 1\qquad 2}\\
\text{Ta được:}\\
\quad I  =\dfrac14\displaystyle\int\limits_1^2u^3du\\
\to I = \dfrac{u^4}{16}\Bigg|_1^2\\
\to I = \dfrac{2^4 - 1}{16}\\
\to I = \dfrac{15}{16}\\
8)\quad I = \displaystyle\int\limits_0^1x^3\sqrt{2-x^2}dx\\
Đặt\,\,u = 2 - x^2\\
\to du = - 2xdx\\
\text{Đổi cận:}\\
x\quad \Big|\quad 0 \qquad 1\\
\overline{u\quad \Big|\quad 2\qquad 1}\\
\text{Ta được:}\\
\quad I = -\dfrac12\displaystyle\int\limits_2^1(2 - u)\sqrt udu\\
\to I = \displaystyle\int\limits_1^2\sqrt udu - \dfrac12\displaystyle\int\limits_1^2\sqrt{u^3}du\\
\to I = \dfrac{2\sqrt{u^3}}{3}\Bigg|_1^2 - \dfrac{\sqrt{u^5}}{5}\Bigg|_1^2\\
\to I = \dfrac{2(\sqrt 8 - 1)}{3} - \dfrac{\sqrt{32} - 1}{5}\\
\to I = \dfrac{8\sqrt2 - 7}{15}\\
9)\quad I = \displaystyle\int\limits_0^1\dfrac{5x}{(x^2 + 4)^2}dx\\
\to I = \dfrac{5}{2}\displaystyle\int\limits_0^1\dfrac{d(x^2 + 4)}{(x^2 + 4)^2}\\
\to I = \dfrac{5}{2}\cdot \left(-\dfrac{1}{x^2 + 4}\right)\Bigg|_0^1\\
\to I = \dfrac{5}{2(0 +4)} - \dfrac{5}{2(1+4)}\\
\to I = \dfrac18\\
10)\quad I  =\displaystyle\int\limits_1^e\dfrac{\sqrt{1 + \ln x}}{x}dx\\
\to I = \displaystyle\int\limits_1^e\sqrt{1 + \ln x}d(1 + \ln x)\\
\to I = \dfrac{2\sqrt{(1+ln x)^3}}{3}\Bigg|_1^e\\
\to I = \dfrac{2[\sqrt{(1 + \ln e)^3} - \sqrt{(1 + \ln 1)^3}]}{3}\\
\to I = \dfrac{4\sqrt2 - 2}{3}\\
11)\quad I = \displaystyle\int\limits_0^{\tfrac{\sqrt2}{2}}\dfrac{x^2dx}{\sqrt{1 - x^2}}\\
Đặt\,\,x = \sin u\\
\to dx = \cos u du\\
\text{Đổi cận:}\\
x\quad \Big|\quad 0 \qquad \dfrac{\sqrt2}{2}\\
\overline{u\quad \Big|\quad 0\qquad \,\,\dfrac{\pi}{4}}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{\sin^2u\cos u}{\sqrt{\cos^2u}}du\\
\to I =  \displaystyle\int\limits_0^{\tfrac{\pi}{4}}(1- \cos^2u)du\\
\to I =  \displaystyle\int\limits_0^{\tfrac{\pi}{4}} du -  \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{1 + \cos2u}{2}du\\
\to I = \dfrac12 \displaystyle\int\limits_0^{\tfrac{\pi}{4}}du - \dfrac12 \displaystyle\int\limits_0^{\tfrac{\pi}{4}}\cos2udu\\
\to I = \dfrac12u\Bigg|_0^{\tfrac{\pi}{4}} - \dfrac14\sin2u\Bigg|_0^{\tfrac{\pi}{4}}\\
\to I = \dfrac{\pi}{8} - \dfrac14\\
12)\quad I =  \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\sin^{2009}x\cos xdx\\
\to I =  \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\sin^{2009}xd(\sin x)\\
\to I = \dfrac{\sin^{2010}x}{2010}\Bigg|_0^{\tfrac{\pi}{2}}\\
\to I = \dfrac{1}{2010}\\
13)\quad I =  \displaystyle\int\limits_\sqrt5^{2\sqrt3}\dfrac{dx}{x\sqrt{x^2+4}}\\
Đặt\,\,u = \sqrt{x^2 + 4}\\
\to du = \dfrac{x}{\sqrt{x^2 + 4}}dx\\
\text{Đổi cận:}\\
x\quad \Big|\quad \sqrt5\qquad 2\sqrt3\\
\overline{u\quad \Big|\quad 3\qquad \quad 4\quad }\\\text{Ta được:}\\
\quad I = \displaystyle\int\limits_3^4\dfrac{1}{u^2 - 4}du\\
\to I = \dfrac14\displaystyle\int\limits_3^4\left(\dfrac{1}{u-2} - \dfrac{1}{u+2}\right)du\\
\to I = \dfrac14\displaystyle\int\limits_3^4\dfrac{d(u-2)}{u-2} - \dfrac14\displaystyle\int\limits_3^4\dfrac{d(u+2)}{u+2}\\
\to I = \dfrac14\ln|u-2|\Bigg|_3^4 - \dfrac14\ln|u+2|\Bigg|_3^4\\
\to I = \dfrac14\ln2 - \dfrac14\ln6 + \dfrac14\ln5\\
\to I = \dfrac14\ln\dfrac{5}{3}\\
14)\quad I = \displaystyle\int\limits_0^1\dfrac{xdx}{\sqrt{2x+1}}\\
Đặt\,\,u = 2x + 1\\
\to du = 2dx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 0 \qquad 1\\
\overline{u\quad \Big|\quad 1\qquad 3}\\
\text{Ta được:}\\
\quad I=\dfrac12 \displaystyle\int\limits_1^3\dfrac{u-1}{2\sqrt u}du\\
\to I = \dfrac14\displaystyle\int\limits_1^3\sqrt u du - \dfrac14\displaystyle\int\limits_1^3\dfrac{1}{\sqrt u}du\\
\to I = \dfrac{\sqrt{u^3}}{6}\Bigg|_1^3 - \dfrac{\sqrt u}{2}\Bigg|_1^3\\
\to I = \dfrac{3\sqrt3 - 1}{6} - \dfrac{\sqrt3 - 1}{2}\\
\to I = \dfrac13\\
15)\quad I = \displaystyle\int\limits_1^4\dfrac{1}{\sqrt{2x+1}}dx\\
Đặt\,\,u = 2x+1\\
\to du = 2dx\\
\text{Đổi cận:}\\
x \quad \Big|\quad 1\qquad 4\\
\overline{u\quad \Big|\quad 3\qquad 9}\\
\text{Ta được:}\\
\quad I = \dfrac12\displaystyle\int\limits_3^9\dfrac{1}{\sqrt u}du\\
\to I = \sqrt u\Bigg|_3^9\\
\to I = 3 - \sqrt3\\
16)\quad I = \displaystyle\int\limits_0^2\left|x^2 - x\right|dx\\
\to I = \displaystyle\int\limits_0^1\left|x^2 - x\right|dx + \displaystyle\int\limits_1^2\left|x^2 - x\right|dx\\
\to I =\displaystyle\int\limits_0^1(x-x^2)dx + \displaystyle\int\limits_1^2(x^2 - x)dx\\
\to I = \displaystyle\int\limits_0^1xdx - \displaystyle\int\limits_0^1x^2dx + \displaystyle\int\limits_1^2x^2dx - \displaystyle\int\limits_1^2xdx\\
\to I = \dfrac{x^2}{2}\Bigg|_0^1 - \dfrac{x^3}{3}\Bigg|_0^1 + \dfrac{x^3}{3}\Bigg|_1^2 - \dfrac{x^2}{2}\Bigg|_1^2\\
\to I = \dfrac12 - \dfrac13 + \dfrac73 - \dfrac32\\
\to I = 1
\end{array}$