Let f(x) = dxx^2/3+2x^1/2.

Question

Let $f(x) = \displaystyle\int\frac{dx}{x^{2/3}+2x^{1/2}}$. If $f(0) = -26+24\log_e(2)$ and $f(1) = a+b\log_e(3)$, $a,b\in\mathbb{Z}$, then $a+b$ is equal to:

Accepted Answer

Let $x=t^6$, $dx=6t^5dt$; $x^{2/3}=t^4$, $x^{1/2}=t^3$:

$$f(x) = \int\frac{6t^5}{t^4+2t^3}dt = \int\frac{6t^2}{t+2}dt = 6\int\left(t-2+\frac{4}{t+2}\right)dt = 3t^2-12t+24\ln|t+2|+C$$

With $t=x^{1/6}$: $f(x) = 3x^{1/3}-12x^{1/6}+24\ln(x^{1/6}+2)+C$. Using $f(0)=-26+24\ln2$: $24\ln2+C = -26+24\ln2 \implies C=-26$.

$$f(1) = 3-12+24\ln3-26 = -35+24\ln3$$

So $a=-35,\,b=24$, and $a+b=-11$.

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