integral of (2x-3)/(sqrt(x^2+2x+5))

해법

\int \frac{2 x - 3}{x ^{2} + 2 x + 5} d x

2 x^{2} + 2 x + 5 - 5 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5) + C

솔루션 단계

\int \frac{2 x - 3}{x ^{2} + 2 x + 5} d x

\frac{2 x - 3}{x ^{2} + 2 x + 5}

확대한다:

\frac{2 x}{x ^{2} + 2 x + 5} - \frac{3}{x ^{2} + 2 x + 5}

= \int \frac{2 x}{x ^{2} + 2 x + 5} - \frac{3}{x ^{2} + 2 x + 5} 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{2 x}{x ^{2} + 2 x + 5} d x - \int \frac{3}{x ^{2} + 2 x + 5} d x

\int \frac{2 x}{x ^{2} + 2 x + 5} d x = 2 (x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5))

\int \frac{3}{x ^{2} + 2 x + 5} d x = 3 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

= 2 (x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)) - 3 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

2 (x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)) - 3 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

간소화하다 :

2 x^{2} + 2 x + 5 - 5 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

= 2 x^{2} + 2 x + 5 - 5 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

솔루션에 상수 추가

= 2 x^{2} + 2 x + 5 - 5 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5) + C