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人気のある微分積分 >

integral from-2 to 2 of 5/((x^2+4)^2)

解

\int_{- 2}^{2} \frac{5}{( x ^{2} + 4 ) ^{2}} d x

解

\frac{5 ( π + 2 )}{32}

+1

十進法表記

0.80337 \dots

解答ステップ

\int_{- 2}^{2} \frac{5}{( x ^{2} + 4 ) ^{2}} d x

定数を除く:

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

= 5 \cdot \int_{- 2}^{2} \frac{1}{( x ^{2} + 4 ) ^{2}} d x

置換積分法を適用する

= 5 \cdot \int_{- 1}^{1} \frac{1}{8 ( u ^{2} + 1 ) ^{2}} d u

定数を除く:

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

= 5 \cdot \frac{1}{8} \cdot \int_{- 1}^{1} \frac{1}{( u ^{2} + 1 ) ^{2}} d u

三角関数による置換を適用する

= 5 \cdot \frac{1}{8} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

三角関数の公式を使用して書き換える

= 5 \cdot \frac{1}{8} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1 + cos ( 2 v )}{2} d v

定数を除く:

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

= 5 \cdot \frac{1}{8} \cdot \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} 1 + cos (2 v) d v

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

= 5 \cdot \frac{1}{8} \cdot \frac{1}{2} (\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d v + \int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 v) d v)

\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d v = \frac{π}{2}

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 v) d v = 1

= 5 \cdot \frac{1}{8} \cdot \frac{1}{2} (\frac{π}{2} + 1)

簡素化

5 \cdot \frac{1}{8} \cdot \frac{1}{2} (\frac{π}{2} + 1) : \frac{5 ( π + 2 )}{32}

= \frac{5 ( π + 2 )}{32}

グラフ

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