【微分積分】4-2-4 基本的な積分公式｜問題集
1.次の積分を求めなさい。
(1)\(\displaystyle \int(3x^{-3}+4x^5)dx\)
\(\displaystyle =3\left(-\frac{1}{2}x^{-2}\right)+\frac{2}{3}x^6+C\)
\(\displaystyle =-\frac{3}{2}x^{-2}+\frac{2}{3}x^6+C\)
\(\displaystyle =-\frac{3}{2}x^{-2}+\frac{2}{3}x^6+C\)
(2)\(\displaystyle \int\left(t^3+\frac{1}{t^2+1}\right)dt\)
\(\displaystyle =\frac{t^4}{4}+\tan^{-1}t+C\)
(3)\(\displaystyle \int-2\tan^2xdx\)
\(\displaystyle =-2\int\frac{1-\cos^2x}{\cos^2x}dx\)
\(\displaystyle =-2\int(\sec^2x-1)dx\)
\(\displaystyle =-2(\tan x-x)+C\)
\(\displaystyle =-2\int(\sec^2x-1)dx\)
\(\displaystyle =-2(\tan x-x)+C\)
(4)\(\displaystyle \int\frac{x^3+1}{\sqrt{x}}dx\)
\(\displaystyle =\int(x^{\frac{5}{2}}+x^{-\frac{1}{2}})dx\)
\(\displaystyle =\frac{2}{7}x^{\frac{7}{2}}+2x^{\frac{1}{2}}+C\)
\(\displaystyle =\frac{2}{7}x^{\frac{7}{2}}+2x^{\frac{1}{2}}+C\)
(5)\(\displaystyle \int\frac{x^2+5}{x^2+4}dx\)
\(\displaystyle =\int\left(1+\frac{1}{x^2+4}\right)dx\)
\(\displaystyle =x+\frac{1}{2}\tan^{-1}\frac{x}{2}+C\)
\(\displaystyle =x+\frac{1}{2}\tan^{-1}\frac{x}{2}+C\)
(6)\(\displaystyle \int\frac{1}{\sqrt{4-t^2}}dt\)
\(\displaystyle =\int\frac{1}{\sqrt{2^2-t^2}}dt\)
\(\displaystyle =\sin^{-1}\frac{t}{2}+C\)
\(\displaystyle =\sin^{-1}\frac{t}{2}+C\)
(7)\(\displaystyle \int\cos^2xdx\)
\(\displaystyle =\int\frac{1+\cos2x}{2}dx\)
\(\displaystyle =\frac{x}{2}+\frac{\sin2x}{4}+C\)
\(\displaystyle =\frac{x}{2}+\frac{\sin2x}{4}+C\)
(8)\(\displaystyle \int\frac{1}{\sqrt{t^2+4}}dt\)
\(\displaystyle =\log|t+\sqrt{t^2+4}|+C\)
(9)\(\displaystyle \int\frac{1}{4-x^2}dx\)
\(\displaystyle =\frac{1}{4}\log\left|\frac{2+x}{2-x}\right|+C\)
次の学習に進もう！