kennisnetto1h

2023-02-18

How to evaluate $\int \sqrt{{x}^{2}+14x}dx$?

affwysaumlgx

We need
${\mathrm{cosh}}^{2}\theta -{\mathrm{sinh}}^{2}\theta =1$
$\mathrm{cosh}2\theta =2{\mathrm{sinh}}^{2}\theta +1$
${\mathrm{sinh}}^{2}\theta =\frac{1}{2}\left(\mathrm{cosh}2\theta -1\right)$
$\mathrm{sinh}2\theta =2\mathrm{sinh}\theta \mathrm{cosh}\theta$
$arc\mathrm{cosh}x=\mathrm{ln}\left(\sqrt{{x}^{2}-1}+x\right)$
By substituting, we perform this integral, but first we simplify.
${x}^{2}+14x={x}^{2}+14x+49-49={\left(x+7\right)}^{2}-49$
The substitution is
$x+7=7\mathrm{cosh}\theta$
$dx=7\mathrm{sinh}\theta d\theta$
$\sqrt{{\left(x+7\right)}^{2}-49}=\sqrt{49{\mathrm{cosh}}^{2}\theta -49}$
$=7\sqrt{{\mathrm{cosh}}^{2}\theta -1}$
$=7\mathrm{sinh}\theta$
${\mathrm{sinh}}^{2}\theta ={\mathrm{cosh}}^{2}\theta -1={\left(\frac{x+7}{7}\right)}^{2}-1$
$=\frac{{x}^{2}+14x+49-49}{49}$
$=\frac{{x}^{2}+14x}{49}$
$\mathrm{sinh}\theta =\frac{1}{7}\sqrt{{x}^{2}+14x}$
Then,$\int \sqrt{{x}^{2}+14x}dx=\int 7\mathrm{sinh}\theta \cdot 7\mathrm{sinh}\theta d\theta$
$=49\int {\mathrm{sinh}}^{2}\theta d\theta$
$=\frac{49}{2}\int \left(\mathrm{cosh}2\theta -1\right)d\theta$
$=\frac{49}{2}\left(\frac{\mathrm{sinh}2\theta }{2}-\theta \right)$
$=\frac{49}{2}\left(\mathrm{sinh}\theta \mathrm{cosh}\theta -arc\mathrm{cosh}\left(\frac{x+7}{7}\right)\right)$
$=\left(\frac{49}{2}\cdot \frac{1}{7}\cdot \sqrt{{x}^{2}+14x}\cdot \frac{x+7}{7}\right)-\frac{49}{2}\left(\mathrm{ln}\left(\sqrt{\frac{{\left(x+7\right)}^{2}}{49}-1}\right)+\frac{x+7}{7}\right)+C$
$=\frac{1}{2}\left(x+7\right)\sqrt{{x}^{2}+14x}-\frac{49}{2}\mathrm{ln}\left(|\frac{1}{7}\left(\sqrt{{x}^{2}+14x}+\left(x+7\right)\right)|\right)+C$

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