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integral of (6x)/(4x^2+4x+10)

해법

\int \frac{6 x}{4 x ^{2} + 4 x + 10} d x

해법

3 (\frac{1}{4} ln 4 x^{2} + 4 x + 10 - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3})) + C

솔루션 단계

\int \frac{6 x}{4 x ^{2} + 4 x + 10} d x

정수를 빼라:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{x}{4 x ^{2} + 4 x + 10} d x

사각형 완성

4 x^{2} + 4 x + 10 : 4 (x + \frac{1}{2})^{2} + 9

= 6 \cdot \int \frac{x}{4 ( x + \frac{1}{2} ) ^{2} + 9} d x

대체 적용

: \int \frac{2 u - 1}{2 ( 4 u ^{2} + 9 )} d u

= 6 \cdot \int \frac{2 u - 1}{2 ( 4 u ^{2} + 9 )} d u

정수를 빼라:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \frac{1}{2} \cdot \int \frac{2 u - 1}{4 u ^{2} + 9} d u

\frac{2 u - 1}{4 u ^{2} + 9}

확대한다:

\frac{2 u}{4 u ^{2} + 9} - \frac{1}{4 u ^{2} + 9}

합계 규칙 적용:

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

= 6 \cdot \frac{1}{2} (\int \frac{2 u}{4 u ^{2} + 9} d u - \int \frac{1}{4 u ^{2} + 9} d u)

\int \frac{2 u}{4 u ^{2} + 9} d u = \frac{1}{4} ln 4 u^{2} + 9

\int \frac{1}{4 u ^{2} + 9} d u = \frac{1}{6} arctan (\frac{2}{3} u)

= 6 \cdot \frac{1}{2} (\frac{1}{4} ln 4 u^{2} + 9 - \frac{1}{6} arctan (\frac{2}{3} u))

뒤로 대체

u = x + \frac{1}{2}

= 6 \cdot \frac{1}{2} (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 9 - \frac{1}{6} arctan (\frac{2}{3} (x + \frac{1}{2})))

6 \cdot \frac{1}{2} (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 9 - \frac{1}{6} arctan (\frac{2}{3} (x + \frac{1}{2})))

간소화하다 :

3 (\frac{1}{4} ln 4 x^{2} + 4 x + 10 - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3}))

= 3 (\frac{1}{4} ln 4 x^{2} + 4 x + 10 - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3}))

솔루션에 상수 추가

= 3 (\frac{1}{4} ln 4 x^{2} + 4 x + 10 - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3})) + C

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