第4章 4.2

(1)

1x2−6x+5\frac{1}{x^{2}-6x+5}

∫1x2−6x+5 dx=∫1(x−5)(x−1) dx=∫{1x−5−1x−1}⋅14 dx=14log⁡∣x−5x−1∣+C\int \frac{1}{x^{2}-6x+5}\,dx=\int \frac{1}{(x-5)(x-1)}\,dx=\int \left\{\frac{1}{x-5}-\frac{1}{x-1}\right\}\cdot\frac{1}{4}\,dx=\frac{1}{4}\log\left|\frac{x-5}{x-1}\right|+C (C は積分定数)

(2)

1x2+4x+5\frac{1}{x^{2}+4x+5}

∫1x2+4x+5 dx=∫1(x+2)2+1 dx=tan⁡−1(x+2)+C\int \frac{1}{x^{2}+4x+5}\,dx=\int \frac{1}{(x+2)^{2}+1}\,dx=\tan^{-1}(x+2)+C (C は積分定数)

(3)

115−4x−4x2\frac{1}{\sqrt{15-4x-4x^{2}}}

∫115−4x−4x2 dx=∫116−(1+2x)2 dx=12sin⁡−11+2x4+C\int \frac{1}{\sqrt{15-4x-4x^{2}}}\,dx=\int \frac{1}{\sqrt{16-(1+2x)^{2}}}\,dx=\frac{1}{2}\sin^{-1}\frac{1+2x}{4}+C (C は積分定数)

(4)

1x2+2x\frac{1}{\sqrt{x^{2}+2x}}

∫1x2+2x dx=∫1(x+1)2−1 dx=log⁡∣(x+1)+(x+1)2−1∣+C=log⁡∣(x+1)+x2+2x∣+C\int \frac{1}{\sqrt{x^{2}+2x}}\,dx=\int \frac{1}{\sqrt{(x+1)^{2}-1}}\,dx=\log\left|(x+1)+\sqrt{(x+1)^{2}-1}\right|+C=\log\left|(x+1)+\sqrt{x^{2}+2x}\right|+C

(C は積分定数)

(5)

14x2+x+1\frac{1}{\sqrt{4x^{2}+x+1}}

∫14x2+x+1 dx=∫1(2x+14)2+1516 dx=12log⁡∣(2x+14)+(2x+14)2+1516∣+C\int \frac{1}{\sqrt{4x^{2}+x+1}}\,dx=\int \frac{1}{\sqrt{\left(2x+\frac{1}{4}\right)^{2}+\frac{15}{16}}}\,dx=\frac{1}{2}\log\left|\left(2x+\frac{1}{4}\right)+\sqrt{\left(2x+\frac{1}{4}\right)^{2}+\frac{15}{16}}\right|+C

=12log⁡∣(2x+14)+4x2+x+1∣+C=\frac{1}{2}\log\left|\left(2x+\frac{1}{4}\right)+\sqrt{4x^{2}+x+1}\right|+C (C は積分定数)

解説: r31bn1z