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人気のある微分積分 >

integral from 0 to 3 of x/(x^2+6x+13)

解

\int_{0}^{3} \frac{x}{x ^{2} + 6 x + 13} d x

解

\frac{ln ( 40 ) - ln ( 13 ) - 3 ( arctan ( 3 ) - arctan ( \frac{3}{2} ) )}{2}

+1

十進法表記

0.16258 \dots

解答ステップ

\int_{0}^{3} \frac{x}{x ^{2} + 6 x + 13} d x

平方完成する

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= \int_{0}^{3} \frac{x}{( x + 3 ) ^{2} + 4} d x

u置換積分法を適用する

= \int_{3}^{6} \frac{u - 3}{u ^{2} + 4} d u

拡張

\frac{u - 3}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{3}{u ^{2} + 4}

総和規則を適用する:

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

= \int_{3}^{6} \frac{u}{u ^{2} + 4} d u - \int_{3}^{6} \frac{3}{u ^{2} + 4} d u

\int_{3}^{6} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (40) - ln (13))

\int_{3}^{6} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (3) - arctan (\frac{3}{2}))

= \frac{1}{2} (ln (40) - ln (13)) - \frac{3}{2} (arctan (3) - arctan (\frac{3}{2}))

簡素化

\frac{1}{2} (ln (40) - ln (13)) - \frac{3}{2} (arctan (3) - arctan (\frac{3}{2})) : \frac{ln ( 40 ) - ln ( 13 ) - 3 ( arctan ( 3 ) - arctan ( \frac{3}{2} ) )}{2}

= \frac{ln ( 40 ) - ln ( 13 ) - 3 ( arctan ( 3 ) - arctan ( \frac{3}{2} ) )}{2}

グラフ

人気の例

integral of 2cos^2(x)tan^3(x)

\int 2 cos^{2} (x) tan^{3} (x) d x

e^{kt}((dy)/(dt)+ky)=0

e^{k t} (\frac{d y}{d t} + k y) = 0

2 x^{2} y y^{^{'}} + y^{2} = 2

tangent (4x-3)^{1/2}

t an g e n t (4 x - 3)^{\frac{1}{2}}

sum from n=2 to infinity of 3/(n*(n+3))

n = 2 \sum \infty \frac{3}{n \cdot ( n + 3 )}