Question

Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.∫ [t / (t⁴ − t² − 2)] dt

Accepted Answer

Start by examining the integral: $$\int \frac{t}{t$$^{4}$$ - t$$^{2}$$ - 2} \, dt. $$Notice that the denominator is a quartic polynomial, which might be factorable to simplify the expression. Try to factor the denominator $$t^{4}$$ - $$t^{2} - 2$. Look for it as a quadratic in terms of t$$^{2}$$: let u$$ = $$t^{2}$$, then rewrite as $$u^{2} - u - 2$. Factor the quadratic in u$: u$$^{2}$$ - u - 2 = (u - 2)(u + 1). $$Substitute back $$u = t$$^{2}$$ to get t$$^{4}$$ - t$$^{2}$$ - 2 = (t$$^{2}$$ - 2)(t$$^{2}$$ + 1). $$Rewrite the integral using the factorization: $$\int \frac{t}{(t$$^{2}$$ - 2)(t$$^{2}$$ + 1)} \, dt. $$Now consider a substitution to simplify the integral. Since the numerator is $$t$$ and the denominator involves $$t^{2}$$ terms, try the substitution $$w = t$$^{2}$$, which implies dw = 2t$$ \, dt$$ or t$$ \, dt = \frac{dw}{2}$$. Rewrite the integral in terms of w$: $$\int \frac{t}{$$(t^{2}$$ - $$2)(t^{2} + 1$$)} \, dt = \int \frac{1}{(w - 2)(w + 1)} \cdot t \, dt = \int \frac{1}{(w - 2)(w + 1)} \cdot \frac{dw}{2} = \frac{1}{2} \int \frac{1}{(w - 2)(w + 1)} \, dw$$. Now the integral is a rational function in w$ that can be solved using partial fraction decomposition.

Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.
∫ [t / (t⁴ − t² − 2)] dt

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Key Concepts

Integration by Substitution

Factoring Polynomials

Partial Fraction Decomposition