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人気のある微分積分 >

integral from 2 to 4 of x/(x^2+6x+13)

解

\int_{2}^{4} \frac{x}{x ^{2} + 6 x + 13} d x

解

\frac{ln ( 53 ) - ln ( 29 ) - 3 ( arctan ( \frac{7}{2} ) - arctan ( \frac{5}{2} ) )}{2}

+1

十進法表記

0.14818 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{5}^{7} \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_{5}^{7} \frac{u}{u ^{2} + 4} d u - \int_{5}^{7} \frac{3}{u ^{2} + 4} d u

\int_{5}^{7} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (53) - ln (29))

\int_{5}^{7} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{7}{2}) - arctan (\frac{5}{2}))

= \frac{1}{2} (ln (53) - ln (29)) - \frac{3}{2} (arctan (\frac{7}{2}) - arctan (\frac{5}{2}))

簡素化

\frac{1}{2} (ln (53) - ln (29)) - \frac{3}{2} (arctan (\frac{7}{2}) - arctan (\frac{5}{2})) : \frac{ln ( 53 ) - ln ( 29 ) - 3 ( arctan ( \frac{7}{2} ) - arctan ( \frac{5}{2} ) )}{2}

= \frac{ln ( 53 ) - ln ( 29 ) - 3 ( arctan ( \frac{7}{2} ) - arctan ( \frac{5}{2} ) )}{2}

グラフ

人気の例

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