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$\int \frac{25 x^2+8}{\sqrt{25 x^2+9}} d x=$
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Verified Answer
The correct answer is:
$\frac{x}{2} \sqrt{25 x^2+9}+\frac{7}{10} \sinh ^{-1}\left(\frac{5 x}{3}\right)+C$
$I=\int \frac{25 x^2+8}{\sqrt{25 x^2+9}}=\int\left[\sqrt{25 x^2+9}-\frac{1}{\sqrt{25 x^2+9}}\right] d x$
$=\frac{x}{2} \sqrt{25 x^2+9}+\frac{9}{2 \times 5} \sinh ^{-1}\left(\frac{5 x}{3}\right)$ $-\frac{1}{5} \sinh ^{-1}\left(\frac{5 x}{3}\right)+C$
$=\frac{x}{2} \sqrt{25 x^2+9}+\frac{7}{10} \sinh ^{-1}\left(\frac{5 x}{3}\right)+C$
$=\frac{x}{2} \sqrt{25 x^2+9}+\frac{9}{2 \times 5} \sinh ^{-1}\left(\frac{5 x}{3}\right)$ $-\frac{1}{5} \sinh ^{-1}\left(\frac{5 x}{3}\right)+C$
$=\frac{x}{2} \sqrt{25 x^2+9}+\frac{7}{10} \sinh ^{-1}\left(\frac{5 x}{3}\right)+C$
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