Question-213342

Tinku Tara November 3, 2024 Integration 0 Comments

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Question Number 213342 by Spillover last updated on 03/Nov/24

Commented by MrGaster last updated on 03/Nov/24

determinant ((( ∫H_x ^(√π) dx=(H_x ^((√π)+1) /( (√π)+1))+C )))

\begin{matrix} \int H_{x}^{\sqrt{π}} d x = \frac{H_{x}^{\sqrt{π} + 1}}{\sqrt{π} + 1} + C \end{matrix}

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