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lim_(x rarr oo)int_(1)^(x)(ln t-(ln^(3)t)/(t))dt

$\lim _{x \rightarrow \infty} \int_{1}^{x}\left(\ln t-\frac{\ln ^{3} t}{t}\right) d t$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\lim _{x \rightarrow \infty} \int_{1}^{x}\left(\ln t-\frac{\ln ^{3} t}{t}\right) d t

Split Integral: First, let's split the integral into two parts. $\newline$ $\int(\ln t - \frac{\ln^3 t}{t}) dt = \int\ln t dt - \int\frac{\ln^3 t}{t} dt$
Antiderivative of $\ln t$ : Now, let's find the antiderivative of $\ln t$ . $\int \ln t \, dt = t \ln t - t + C$
Antiderivative of $(\ln^3 t)/t$ : Next, we need to find the antiderivative of $(\ln^3 t)/t$ . Let $u = \ln t$ , then $du = dt/t$ . \int(\ln^\(3 t)/t \, dt = \int u^ $3$ \, du = (u^ $4$ )/ $4$ + C = (\ln^ $4$ t)/ $4$ + C
Combine Antiderivatives: Combine the antiderivatives. $\int(\ln t - \frac{\ln^3 t}{t}) dt = (t \ln t - t) - \frac{\ln^4 t}{4} + C$
Evaluate Definite Integral: Now, let's evaluate the definite integral from $1$ to $x$ . $\int_{1}^{x}(\ln t - \frac{\ln^3 t}{t}) dt = \left[(x \ln x - x) - \frac{\ln^4 x}{4}\right] - \left[(1 \ln 1 - 1) - \frac{\ln^4 1}{4}\right]$
Simplify Expression: Simplify the expression, noting that $\ln 1 = 0$ . $\int_{1}^{x}(\ln t - \frac{\ln^3 t}{t}) dt = (x \ln x - x) - \frac{\ln^4 x}{4} + 1$
Take Limit: Now, take the limit as $x$ approaches infinity. $\lim_{x \to \infty}(x \ln x - x - (\ln^4 x)/4 + 1)$
Take Limit: Now, take the limit as $x$ approaches infinity. $\lim_{x \to \infty}(x \ln x - x - (\ln^4 x)/4 + 1)$ As $x$ approaches infinity, $x \ln x$ grows without bound, and so does $-x$ . However, the term $(\ln^4 x)/4$ grows much slower than $x \ln x$ and $-x$ , so it becomes negligible. $\lim_{x \to \infty}(x \ln x - x - (\ln^4 x)/4 + 1) = \infty - \infty - 0 + 1$

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