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人気のある微分積分 >

integral of 1/(xsqrt(1+ln(x^2)))

解

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

解

1 + 2 ln (x) + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{1}{2 u} 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 \frac{1}{u} d u

べき乗則を適用する

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

代用を戻す

u = 1 + ln (x^{2})

= \frac{1}{2} \cdot 2 (1 + ln (x^{2}))^{\frac{1}{2}}

簡素化

\frac{1}{2} \cdot 2 (1 + ln (x^{2}))^{\frac{1}{2}} : 1 + 2 ln (x)

= 1 + 2 ln (x)

定数を解答に追加する

= 1 + 2 ln (x) + C

グラフ

人気の例

(\partial)/(\partial x)(sqrt(25-x^2-5y))

\frac{\partial}{\partial x} (25 - x^{2} - 5 y)

sum from n=0 to infinity of 5/(pi^n)

n = 0 \sum \infty \frac{5}{π ^{n}}

derivative of 4+2x-3x^2-5x^3-8x^4+9x^5

\frac{d}{d x} (4 + 2 x - 3 x^{2} - 5 x^{3} - 8 x^{4} + 9 x^{5})

integral from 0 to 2pi of sqrt(20+16t^2)

\int_{0}^{2 π} 20 + 16 t^{2} d t

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x \to - 2 lim ((x + 1)^{4} (2 x^{2}))