Question

If $I=\int \frac{dt}{t^{\frac{4}{3}}+t}$, then which of the following is/are equal to $I$:
(where $C$ is integration constant)

• A. $3\ln |1+t^{\frac{1}{3}}|+\frac{1}{3}\ln |t|+C$
• B. $-3\ln |1+t^{-\frac{1}{3}}|+C$
• C. $3\ln |1+t^{-\frac{1}{3}}|+C$
• D. $-3\ln |1+t^{\frac{1}{3}}|+\ln |t|+C$

Answer 1

Answer: (B), (D)

$-3\ln |1+t^{\frac{1}{3}}|+\ln |t|+C$

$I=\int \frac{dt}{t^{\frac{4}{3}}+t}=\int \frac{dt}{t^{\frac{4}{3}}(1+t^{\frac{-1}{3}})}$
Put $1+t^{\frac{-1}{3}}=x$
$\Rightarrow -\frac{1}{3}{t}^{\frac{-4}{3}}dt=dx\Rightarrow \frac{dt}{t^{\frac{4}{3}}}=-3dx$
$\Rightarrow I=-\int \frac{3}{x}dx=-3\ln |x|+C\Rightarrow I=-3\ln |1+t^{\frac{-1}{3}}|+C\Rightarrow I=-3\ln |1+t^{\frac{1}{3}}|+\ln |t|+C$