integral of (x-1)/(sqrt(x^2+2x+2))

해법

\int \frac{x - 1}{x ^{2} + 2 x + 2} d x

x^{2} + 2 x + 2 - 2 ln x^{2} + 2 x + 2 + x + 1 + C

솔루션 단계

\int \frac{x - 1}{x ^{2} + 2 x + 2} d x

\frac{x - 1}{x ^{2} + 2 x + 2}

확대한다:

\frac{x}{x ^{2} + 2 x + 2} - \frac{1}{x ^{2} + 2 x + 2}

= \int \frac{x}{x ^{2} + 2 x + 2} - \frac{1}{x ^{2} + 2 x + 2} d x

합계 규칙 적용:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{x ^{2} + 2 x + 2} d x - \int \frac{1}{x ^{2} + 2 x + 2} d x

\int \frac{x}{x ^{2} + 2 x + 2} d x = x^{2} + 2 x + 2 - ln x^{2} + 2 x + 2 + x + 1

\int \frac{1}{x ^{2} + 2 x + 2} d x = ln x^{2} + 2 x + 2 + x + 1

= x^{2} + 2 x + 2 - ln x^{2} + 2 x + 2 + x + 1 - ln x^{2} + 2 x + 2 + x + 1

단순화

= x^{2} + 2 x + 2 - 2 ln x^{2} + 2 x + 2 + x + 1

솔루션에 상수 추가

= x^{2} + 2 x + 2 - 2 ln x^{2} + 2 x + 2 + x + 1 + C