Question

`\int\limits^9_3 {(1)/((x+21)√(x+22) ) } \, dx`

Answer 1

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

`\displaystyle \int_3^9 (\mathrm dx)/((x+21)√(x+22)) = \int_(25)^(31) (\mathrm dy)/((y-1)√(y))`

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

`\displaystyle \int_3^9 (\mathrm dx)/((x+21)√(x+22)) = \int_(√(25))^(√(31)) (2z)/((z^2-1)z) \\\\ = \int_5^(√(31)) (2)/(z^2-1)\,\mathrm dz`

Expand the integrand into partial fractions:

`(2)/(z^2-1) = \frac1{z-1}-\frac1{z+1}`

Then we have

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