The value of the integral I=∫0∞dxx2+x is equal to aπbb, (b is prime) then a + b is equal to _______

Consider x=t2 and solve the integral∫0∞2t dtt4+t=2∫0∞dt1+t3=2[13∫0∞dt1+t+13∫0∞−t+2t2−t+1dt] =2[13ln(1+t)|0∞−16ln(t2−t+1)|0∞]+12∫0∞dt(t−12)2+(34) =23[π2−(−π6)]=4π33

The value of the integral I=∫0∞dxx2+x is equal to aπbb, (b is prime) then a + b is equal to _______

Consider x=t2 and solve the integral∫0∞2t dtt4+t=2∫0∞dt1+t3=2[13∫0∞dt1+t+13∫0∞−t+2t2−t+1dt] =2[13ln(1+t)|0∞−16ln(t2−t+1)|0∞]+12∫0∞dt(t−12)2+(34) =23[π2−(−π6)]=4π33

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The value of the integral $I = \int_{0}^{\infty} \frac{d x}{x^{2} + \sqrt{x}}$ is equal to $\frac{a π}{b \sqrt{b}}$ , (b is prime) then a + b is equal to _______

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Detailed Solution

Consider $x = t^{2}$ and solve the integral

$\int_{0}^{\infty} \frac{2 t d t}{t^{4} + t} = 2 \int_{0}^{\infty} \frac{d t}{1 + t^{3}} = 2 [\frac{1}{3} \int_{0}^{\infty} \frac{d t}{1 + t} + \frac{1}{3} \int_{0}^{\infty} \frac{- t + 2}{t^{2} - t + 1} d t] = 2 [\frac{1}{3} \ln (1 + t) |_{0}^{\infty} - \frac{1}{6} \ln (t^{2} - t + 1) |_{0}^{\infty}] + \frac{1}{2} \int_{0}^{\infty} \frac{d t}{{(t - \frac{1}{2})}^{2} + (\frac{3}{4})} = \frac{2}{\sqrt{3}} [\frac{π}{2} - (\frac{- π}{6})] = \frac{4 π}{3 \sqrt{3}}$