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int(te^(t))/(1+e^(2t))dt

$\int \frac{t e^{t}}{1+e^{2 t}} d t$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\int \frac{t e^{t}}{1+e^{2 t}} d t

Rewrite denominator identity: We start by noticing that the denominator can be rewritten using a hyperbolic function identity, $\cosh^2(x) = \frac{e^{2x} + 1}{2}$ . Thus, $1 + e^{2t} = 2\cosh^2(t)$ . Rewrite the integral: $\int \frac{te^{t}}{1+e^{2t}}\,dt = \int \frac{te^{t}}{2\cosh^2(t)}\,dt$
Rewrite integral: Next, let's use a substitution. Let $u = t$ , then $du = dt$ . This doesn't simplify the integral directly, but it helps to reframe it: $\newline$ $\int\frac{u e^u}{2\cosh^2(u)} \, du$
Use substitution: We can try another substitution: let $v = e^t$ , then $dv = e^t dt$ , or $dt = \frac{dv}{v}$ . Substituting these into the integral: $\newline$ $\int\frac{t v}{2\cosh^2(t)} \cdot \frac{dv}{v}$ $\newline$ This simplifies to: $\newline$ $\int\frac{t}{2\cosh^2(t)} dv$

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