Интегралом помогите, прошу. Created by assaper. matematika-ru

Интегралом помогите, прошу...

igundane

$\int \dfrac{dx}{\sqrt{1-25x^2}\cdot arcsin5x}=\int \dfrac{1}{arcsin5x}\cdot \dfrac{dx}{\sqrt{1-25x^2}}=\\\\\\=\Big[\ t=arcsin5x\ ,\ dt=\dfrac{5\, dx}{\sqrt{1-25x^2}}\ \Big]=\dfrac{1}{5}\int \dfrac{1}{t}\cdot dt=\dfrac{1}{5}\int \dfrac{dt}{t}=\dfrac{1}{5}\cdot ln|t|+C=\\\\\\=\dfrac{1}{5}\cdot ln\, |arcsin5x|+C\ ;$

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assaper Там 1/5ln+c?

NNNLLL54 нет

NNNLLL54 dx = ?

NNNLLL54 исправьте выражение для dx ...

NNNLLL54

Verified answer

$\int \frac{dx}{\sqrt{1-25x^2}arcsin5x}=\begin{vmatrix}arcsin5x=t\\ dx=\frac{\sqrt{1-25x^2}}{5}dt\end{vmatrix}=\frac{1}{5}\int \frac{dt}{t}=\frac{1}{5}lnt+C\Rightarrow \frac{1}{5}ln\left ( arcsin5x \right )+C$

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