Question
Let $f(x) = \int_{}^{}\ \frac{dx}{x^{\left( \frac{2}{3} \right)} + 2x^{\left( \frac{1}{2} \right)}}$ be such that $f(0) = - 26 + 24\log_{e}(2)$. If $f(1) = a + b\log_{e}(3)$, where $a,b \in \mathbf{Z}$, then $a + b$ is equal to:
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Solution
$f(x) = \int_{}^{}\ \frac{dx}{x^{2/3} + 2x^{1/2}}$
Put $x = t^{6} \Rightarrow dx = 6t^{5}dt$
$${= \int\frac{6t^{5}dt}{t^{4} + 2t^{3}} = 6\int\frac{\left( t^{2} - 4 \right) + 4}{t + 2}dt }{= 6\left\lbrack \int(t - 2)dt + 4\int\frac{1}{t + 2}dt \right\rbrack }{= 6\left\lbrack \frac{t^{2}}{2} - 2t + 4\mathcal{l}n(t + 2) \right\rbrack + C }{= 3x^{1/3} - 12x^{1/6} + 24\mathcal{l}n\left( x^{1/6} + 2 \right) + C }$$$f(0) = 24\mathcal{l}n2 + C = - 26 + 24\mathcal{l}n2$ (given)
$$\Rightarrow C = - 26 $$Now
$f(1) = - 35 + 24\mathcal{l}n3 = a + b\mathcal{l}n3$ (as given in ques.)
$${\Rightarrow a = - 35\&\text{ }b = 24 }{\Rightarrow a + b = - 11}$$
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