与えられた不定積分を計算します。 (1) $\int \frac{\sqrt[5]{x}+2x}{x^2} dx$ (2) $\int (x^3\sqrt{x}-5^x) dx$ (3) $\int (4e^x+3\tan{x}) dx$ (4) $\int (2x+1)(x^2+x+1) dx$ (5) $\int \frac{\log{|x|}}{x} dx$ (6) $\int \frac{2x+3}{x^2+3x+1} dx$ (7) $\int x\cos{x} dx$ (8) $\int \frac{1}{x^2+x-6} dx$ (9) $\int \frac{1}{x^2(x+1)} dx$

はい、承知いたしました。それでは、画像にある問題のうち、(1), (2), (3), (4), (5), (6), (7), (8), (9) の不定積分を求めます。

1. 問題の内容

与えられた不定積分を計算します。

(1) $\int \frac{\sqrt[5]{x}+2x}{x^2} dx$

(2) $\int (x^3\sqrt{x}-5^x) dx$

(3) $\int (4e^x+3\tan{x}) dx$

(4) $\int (2x+1)(x^2+x+1) dx$

(5) $\int \frac{\log{|x|}}{x} dx$

(6) $\int \frac{2x+3}{x^2+3x+1} dx$

(7) $\int x\cos{x} dx$

(8) $\int \frac{1}{x^2+x-6} dx$

(9) $\int \frac{1}{x^2(x+1)} dx$

2. 解き方の手順

(1) $\int \frac{\sqrt[5]{x}+2x}{x^2} dx = \int (x^{-9/5} + 2x^{-1}) dx = \frac{x^{-4/5}}{-4/5} + 2\log{|x|} + C = -\frac{5}{4}x^{-4/5} + 2\log{|x|} + C$

(2) $\int (x^3\sqrt{x}-5^x) dx = \int (x^{7/2}-5^x) dx = \frac{x^{9/2}}{9/2} - \frac{5^x}{\log{5}} + C = \frac{2}{9}x^{9/2} - \frac{5^x}{\log{5}} + C$

(3) $\int (4e^x+3\tan{x}) dx = 4\int e^x dx + 3\int \tan{x} dx = 4e^x - 3\log{|\cos{x}|} + C$

(4) $\int (2x+1)(x^2+x+1) dx = \int (2x^3 + 3x^2 + 3x + 1) dx = \frac{2x^4}{4} + \frac{3x^3}{3} + \frac{3x^2}{2} + x + C = \frac{1}{2}x^4 + x^3 + \frac{3}{2}x^2 + x + C$

(5) $\int \frac{\log{|x|}}{x} dx$.

$u = \log{|x|}$ とおくと、$du = \frac{1}{x} dx$.

$\int u du = \frac{1}{2}u^2 + C = \frac{1}{2}(\log{|x|})^2 + C$

(6) $\int \frac{2x+3}{x^2+3x+1} dx$.

$u = x^2+3x+1$ とおくと、$du = (2x+3)dx$.

$\int \frac{1}{u} du = \log{|u|} + C = \log{|x^2+3x+1|} + C$

(7) $\int x\cos{x} dx$.

部分積分を使用する。$u=x, dv=\cos{x} dx$.

$du=dx, v=\sin{x}$.

$\int x\cos{x} dx = x\sin{x} - \int \sin{x} dx = x\sin{x} + \cos{x} + C$

(8) $\int \frac{1}{x^2+x-6} dx = \int \frac{1}{(x+3)(x-2)} dx = \int \frac{A}{x+3} + \frac{B}{x-2} dx$

$1 = A(x-2) + B(x+3)$

$x=2 \implies 1 = 5B \implies B = 1/5$

$x=-3 \implies 1 = -5A \implies A = -1/5$

$\int \frac{-1/5}{x+3} + \frac{1/5}{x-2} dx = -\frac{1}{5}\log{|x+3|} + \frac{1}{5}\log{|x-2|} + C = \frac{1}{5}\log{\left|\frac{x-2}{x+3}\right|} + C$

(9) $\int \frac{1}{x^2(x+1)} dx = \int \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} dx$

$1 = Ax(x+1) + B(x+1) + Cx^2$

$x=0 \implies 1 = B$

$x=-1 \implies 1 = C$

$x=1 \implies 1 = 2A + 2B + C = 2A + 2 + 1 \implies -2 = 2A \implies A = -1$

$\int \frac{-1}{x} + \frac{1}{x^2} + \frac{1}{x+1} dx = -\log{|x|} - \frac{1}{x} + \log{|x+1|} + C = \log{\left|\frac{x+1}{x}\right|} - \frac{1}{x} + C$

3. 最終的な答え

(1) $-\frac{5}{4}x^{-4/5} + 2\log{|x|} + C$

(2) $\frac{2}{9}x^{9/2} - \frac{5^x}{\log{5}} + C$

(3) $4e^x - 3\log{|\cos{x}|} + C$

(4) $\frac{1}{2}x^4 + x^3 + \frac{3}{2}x^2 + x + C$

(5) $\frac{1}{2}(\log{|x|})^2 + C$

(6) $\log{|x^2+3x+1|} + C$

(7) $x\sin{x} + \cos{x} + C$

(8) $\frac{1}{5}\log{\left|\frac{x-2}{x+3}\right|} + C$

(9) $\log{\left|\frac{x+1}{x}\right|} - \frac{1}{x} + C$