Integration of a Substitute

Calculus Level 2

If $f\left( \frac{3x-4}{3x+4} \right) = x + 2$ , then what is the antiderivative of $f(x)$ ?

Clarification: Take $C$ as an arbitrary constant.

e^{x+2} \ln(\frac{3x - 4}{3x + 4}) + C

e^{x-2} \ln(\frac{3x - 4}{3x + 4}) + C

\frac{-8}{3}\ln(x-1) + \frac{2x}{3} + C

None of these choices

e^{\frac{3x - 4}{3x + 4}} - \frac{x^2}{2} + C

\frac{8}{3}\ln(x-1) + \frac{2x}{3} + C

e^{\frac{3x + 4}{3x - 4}} - \frac{x^2}{2} + C

1 solution

Vishwak Srinivasan
Jul 3, 2015

What matters the most here is the conversion of $f(\dfrac{3x-4}{3x+4})$ to $f(x)$

So let us take $t = \dfrac{3x-4}{3x+4}$

$3xt + 4t = 3x - 4 \Rightarrow \dfrac{4t + 4}{3-3t} = x$

$\Rightarrow f(t) = \dfrac{4t+4}{3-3t} + 2 = \dfrac{10 - 2t}{3 - 3t} = \dfrac{2t - 10}{3t - 3}$

$\Rightarrow f(x) = \dfrac{2x - 10}{3x - 3}$

Now that the transformation is complete, the anti-derivative is nothing but the integral

$\displaystyle \int f(x) dx = \int \dfrac{2x - 10}{3x - 3} dx = \int \dfrac{2x-2}{3x-3} dx + \int \dfrac{-8}{3x-3} dx$

$\int f(x) dx = \dfrac{2x}{3} + c_1 + \dfrac{-8}{3}\ln(x-1) + c_2$

$\int f(x) dx = \dfrac{2x}{3} + \dfrac{-8}{3}\ln(x-1) + C$

Nice question :)

saharsh rathi - 4 years, 8 months ago

Nice method of solving....!!!

Vishal Dbhat - 3 years, 6 months ago

Very nice method of solving

Vasant Halpet - 2 years, 1 month ago