(1) $\int (-2) dx$ (2) $\int 5x^2 dx$ (3) $\int (2x-3) dx$ (4) $\int (x^2 - 4x + 3) dx$ (5) $\int (-2x^3 - 6x^2 - 2x + 5) dx$ (6) $\int (10x^4 - 9x^2 + 3) dx$ (7) $\int (3x+2)(3x-2) dx$ (8) $\int t(t+5) dt$ (9) $\int (u+1)(u-2) du$ (10) $\int \frac{1}{x^2} dx$ (11) $\int \frac{1}{\sqrt{x}} dx$ (12) $\int (1+2\sin x) dx$ (13) $\int (3\sin x - 4\cos x) dx$ (14) $\int (2e^x - x^2 + 3) dx$

はい、承知いたしました。それでは、画像に写っている数学の問題を解きます。

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1. 問題の内容**

1. 次の不定積分を求めよ。

(1) $\int (-2) dx$

(2) $\int 5x^2 dx$

(3) $\int (2x-3) dx$

(4) $\int (x^2 - 4x + 3) dx$

(5) $\int (-2x^3 - 6x^2 - 2x + 5) dx$

(6) $\int (10x^4 - 9x^2 + 3) dx$

(7) $\int (3x+2)(3x-2) dx$

(8) $\int t(t+5) dt$

(9) $\int (u+1)(u-2) du$

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

(11) $\int \frac{1}{\sqrt{x}} dx$

(12) $\int (1+2\sin x) dx$

(13) $\int (3\sin x - 4\cos x) dx$

(14) $\int (2e^x - x^2 + 3) dx$

2. 次の不定積分を求めよ。

(1) $\int x \cdot \sin x dx$

(2) $\int x \cdot e^x dx$

(3) $\int x \cdot \log x dx$

(4) $\int x \cdot e^{-\lambda x} dx$ (ただし、$\lambda > 0$ の定数)

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2. 解き方の手順**

1. の不定積分を求めます。

(1) $\int (-2) dx = -2x + C$

(2) $\int 5x^2 dx = 5 \cdot \frac{x^3}{3} + C = \frac{5}{3}x^3 + C$

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

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

(5) $\int (-2x^3 - 6x^2 - 2x + 5) dx = -2 \cdot \frac{x^4}{4} - 6 \cdot \frac{x^3}{3} - 2 \cdot \frac{x^2}{2} + 5x + C = -\frac{1}{2}x^4 - 2x^3 - x^2 + 5x + C$

(6) $\int (10x^4 - 9x^2 + 3) dx = 10 \cdot \frac{x^5}{5} - 9 \cdot \frac{x^3}{3} + 3x + C = 2x^5 - 3x^3 + 3x + C$

(7) $\int (3x+2)(3x-2) dx = \int (9x^2 - 4) dx = 9 \cdot \frac{x^3}{3} - 4x + C = 3x^3 - 4x + C$

(8) $\int t(t+5) dt = \int (t^2 + 5t) dt = \frac{t^3}{3} + 5 \cdot \frac{t^2}{2} + C = \frac{1}{3}t^3 + \frac{5}{2}t^2 + C$

(9) $\int (u+1)(u-2) du = \int (u^2 - u - 2) du = \frac{u^3}{3} - \frac{u^2}{2} - 2u + C$

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

(11) $\int \frac{1}{\sqrt{x}} dx = \int x^{-\frac{1}{2}} dx = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = 2\sqrt{x} + C$

(12) $\int (1+2\sin x) dx = x - 2\cos x + C$

(13) $\int (3\sin x - 4\cos x) dx = -3\cos x - 4\sin x + C$

(14) $\int (2e^x - x^2 + 3) dx = 2e^x - \frac{x^3}{3} + 3x + C$

2. の不定積分を求めます。部分積分を使います。

(1) $\int x \cdot \sin x dx$

$u = x$, $dv = \sin x dx$

$du = dx$, $v = -\cos x$

$\int x \sin x dx = -x\cos x - \int (-\cos x) dx = -x\cos x + \int \cos x dx = -x\cos x + \sin x + C$

(2) $\int x \cdot e^x dx$

$u = x$, $dv = e^x dx$

$du = dx$, $v = e^x$

$\int x e^x dx = xe^x - \int e^x dx = xe^x - e^x + C$

(3) $\int x \cdot \log x dx$

$u = \log x$, $dv = x dx$

$du = \frac{1}{x} dx$, $v = \frac{x^2}{2}$

$\int x \log x dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \log x - \int \frac{x}{2} dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + C$

(4) $\int x \cdot e^{-\lambda x} dx$

$u = x$, $dv = e^{-\lambda x} dx$

$du = dx$, $v = -\frac{1}{\lambda}e^{-\lambda x}$

$\int x e^{-\lambda x} dx = -\frac{x}{\lambda}e^{-\lambda x} - \int -\frac{1}{\lambda}e^{-\lambda x} dx = -\frac{x}{\lambda}e^{-\lambda x} + \frac{1}{\lambda} \int e^{-\lambda x} dx = -\frac{x}{\lambda}e^{-\lambda x} + \frac{1}{\lambda} \cdot (-\frac{1}{\lambda} e^{-\lambda x}) + C = -\frac{x}{\lambda}e^{-\lambda x} - \frac{1}{\lambda^2} e^{-\lambda x} + C = -\frac{1}{\lambda}e^{-\lambda x}(x+\frac{1}{\lambda}) + C$

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3. 最終的な答え**

1. (1) $-2x + C$

(2) $\frac{5}{3}x^3 + C$

(3) $x^2 - 3x + C$

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

(5) $-\frac{1}{2}x^4 - 2x^3 - x^2 + 5x + C$

(6) $2x^5 - 3x^3 + 3x + C$

(7) $3x^3 - 4x + C$

(8) $\frac{1}{3}t^3 + \frac{5}{2}t^2 + C$

(9) $\frac{u^3}{3} - \frac{u^2}{2} - 2u + C$

(10) $-\frac{1}{x} + C$

(11) $2\sqrt{x} + C$

(12) $x - 2\cos x + C$

(13) $-3\cos x - 4\sin x + C$

(14) $2e^x - \frac{x^3}{3} + 3x + C$

2. (1) $-x\cos x + \sin x + C$

(2) $xe^x - e^x + C$

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

(4) $-\frac{1}{\lambda}e^{-\lambda x}(x+\frac{1}{\lambda}) + C$