int frac{25 x^2+8}{sqrt{25 x^2+9}} d x= | vedclass

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Question

(C) Let $I = \int \frac{25 x^2+8}{\sqrt{25 x^2+9}} dx$. We can rewrite the numerator as $(25x^2 + 9) - 1$. So, $I = \int \frac{25x^2 + 9 - 1}{\sqrt{25

Vedclass · Accepted Answer

$\frac{x}{2} \sqrt{25 x^2+9}+\frac{7}{10} \sinh ^{-1}\left(\frac{5 x}{3}\right)+C$

Vedclass · Answer

(C) Let $I = \int \frac{25 x^2+8}{\sqrt{25 x^2+9}} dx$. We can rewrite the numerator as $(25x^2 + 9) - 1$. So, $I = \int \frac{25x^2 + 9 - 1}{\sqrt{25x^2 + 9}} dx = \int \sqrt{25x^2 + 9} dx - \int \frac{1}{\sqrt{25x^2 + 9}} dx$. Using the standard integral formula $\int \sqrt{a^2x^2 + b^2} dx = \frac{x}{2}\sqrt{a^2x^2 + b^2} + \frac{b^2}{2a}\sinh^{-1}(\frac{ax}{b})$ and $\int \frac{1}{\sqrt{a^2x^2 + b^2}} dx = \frac{1}{a}\sinh^{-1}(\frac{ax}{b})$. Here $a=5$ and $b=3$. $I = [\frac{x}{2}\sqrt{25x^2 + 9} + \frac{9}{2(5)}\sinh^{-1}(\frac{5x}{3})] - \frac{1}{5}\sinh^{-1}(\frac{5x}{3}) + C$. $I = \frac{x}{2}\sqrt{25x^2 + 9} + \frac{9}{10}\sinh^{-1}(\frac{5x}{3}) - \frac{2}{10}\sinh^{-1}(\frac{5x}{3}) + C$. $I = \frac{x}{2}\sqrt{25x^2 + 9} + \frac{7}{10}\sinh^{-1}(\frac{5x}{3}) + C$.

$\int \frac{25 x^2+8}{\sqrt{25 x^2+9}} d x=$

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