Let [ f(x)= d xx^2 / 3+2 x^1 / 2, ] be such that f(0) = -26 + 24 _e(2).

Question

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

Accepted Answer

Given

$f(x)=\int \frac{dx}{{x}^{2/3}+2{x}^{1/2}},$

$f(0)=-26+24\ln 2$

$f(1)=a+b\ln 3,\, a,b\in Z$

Use substitution Let $x={t}^{6}\implies dx=6{t}^{5}dt$

$\begin{gathered} {x}^{2/3}={({t}^{6})}^{2/3}={t}^{4} \ {x}^{1/2}={({t}^{6})}^{1/2}={t}^{3} \end{gathered}$

Substituting these into the integral :

$f(x)=\int \frac{6{t}^{5}}{{t}^{4}+2{t}^{3}}dt=\int \frac{6{t}^{5}}{{t}^{3}(t+2)}dt=\int \frac{6{t}^{2}}{t+2}dt$

manipulate

$\frac{{t}^{2}}{t+2}=\frac{({t}^{2}-4)+4}{t+2}=\frac{(t-2)(t+2)+4}{t+2}=(t-2)+\frac{4}{t+2}$

Now, integrate term by term:

$\begin{gathered} f(x)=6\int (t-2+\frac{4}{t+2})dt \ f(x)=6[\frac{{t}^{2}}{2}-2t+4\ln |t+2|]+C \end{gathered}$

$f(x)=3{t}^{2}-12t+24\ln |t+2|+C$

Back-substitute $t={x}^{1/6}$ :

$f(x)=3{x}^{1/3}-12{x}^{1/6}+24\ln |{x}^{1/6}+2|+C$

Finding the Constant $C$

We are given the initial condition $f(0)=-26+24\ln (2)$ :

$\begin{gathered} f(0)=3(0)-12(0)+24\ln (0+2)+C \ -26+24\ln (2)=24\ln (2)+C\implies C=-26 \end{gathered}$

So, the complete function is:

$f(x)=3{x}^{1/3}-12{x}^{1/6}+24\ln ({x}^{1/6}+2)-26$

Solving for $a$ and $b$

Now we evaluate $f(1)$ :

$\begin{gathered} f(1)=3(1{)}^{1/3}-12(1{)}^{1/6}+24\ln ({1}^{1/6}+2)-26 \ f(1)=3-12+24\ln (3)-26 \ f(1)=-35+24\ln (3) \end{gathered}$

Comparing this to the given form $f(1)=a+b\ln (3)$ :

$a=-35\, \, b=24$

Then

$a+b=-35+24=-11$

Solutions · 1