ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral of ln(1+r^2)r

解

\int ln (1 + r^{2}) r d r

解

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

解答ステップ

\int ln (1 + r^{2}) r d r

u置換積分法を適用する

= \int \frac{ln ( u )}{2} d u

定数を除く:

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

= \frac{1}{2} \cdot \int ln (u) d u

部分積分を適用する

= \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} (u ln (u) - u)

代用を戻す

u = 1 + r^{2}

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

拡張

(1 + r^{2}) ln (1 + r^{2}) - (1 + r^{2}) : ln (1 + r^{2}) + r^{2} ln (1 + r^{2}) - 1 - r^{2}

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

定数を解答に追加する

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

グラフ

人気の例

(\partial)/(\partial x)(yzln(xy))

\frac{\partial}{\partial x} (yz ln (x y))

y(x+y+1)dx+(x+2y)dy=0

y (x + y + 1) d x + (x + 2 y) d y = 0

x^{''}-2x-5=0,x(0)=1,x(1)=3

x^{^{''}} - 2 x - 5 = 0, x (0) = 1, x (1) = 3

integral of (39)/((1-x^2)^{3/2)}

\int \frac{39}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

limit as x approaches 2 of 2x-3

x \to 2 lim (2 x - 3)