Question

$\int\limits^9_3 {\frac{1}{(x+21)\sqrt{x+22} } } \, dx$

Answer 1

Let y = x + 22 and dy = dx, so the integral becomes

$\displaystyle \int_3^9 \frac{\mathrm dx}{(x+21)\sqrt{x+22}} = \int_{25}^{31} \frac{\mathrm dy}{(y-1)\sqrt{y}}$

Now let z = √y, so that z ² = y. Then 2z dz = dy, and the integral becomes

$\displaystyle \int_3^9 \frac{\mathrm dx}{(x+21)\sqrt{x+22}} = \int_{\sqrt{25}}^{\sqrt{31}} \frac{2z}{(z^2-1)z} \\\\ = \int_5^{\sqrt{31}} \frac{2}{z^2-1}\,\mathrm dz$

Expand the integrand into partial fractions:

$\dfrac{2}{z^2-1} = \dfrac1{z-1}-\dfrac1{z+1}$

Then we have

$\displaystyle \int_3^9 \frac{\mathrm dx}{(x+21)\sqrt{x+22}} = \int_5^{\sqrt{31}}\left(\frac1{z-1}-\frac1{z+1}\right)\,\mathrm dz \\\\ = \left(\ln|z-1|-\ln|z+1|\right)\bigg|_5^{\sqrt{31}} \\\\ =\left[\ln\left|\frac{z-1}{z+1}\right|\right]\bigg|_5^{\sqrt{31}} \\\\ =\ln\left(\frac{\sqrt{31}-1}{\sqrt{31}+1}\right) - \ln\left(\frac{4}{6}\right) \\\\ =\ln\left(32-2\sqrt{31}\right) - \ln\left(\frac23\right) \\\\ =\boxed{\ln\left(48-3\sqrt{31}\right)}$