Question

Evaluate the integrals in Exercises 31–78.77. ∫dt/((t+1)√(t²+2t-8))

Accepted Answer

Start by examining the integral: $$\int \frac{dt}{(t+1) \sqrt{t$$^{2}$$ + 2t - 8}}. $$Notice the expression under the square root, $$t^{2} + 2t - 8$, which can be simplified by completing the square. Complete the square for the quadratic inside the square root: t$$^{2}$$ + 2t - 8 = (t$$^{2}$$ + 2t + 1) - 1 - 8 = (t + 1$$)^{2}$$ - 9. $$Rewrite the integral as $$\int \frac{dt}{(t+1) \sqrt{(t+1$$)^{2}$$ - 9}}. $$Make the substitution $$x = t + 1$$, so that $$dt = dx. $$The integral becomes $$\int \frac{dx}{x \sqrt{x$$^{2}$$ - 9}}. $$This substitution simplifies the integral and prepares it for a trigonometric substitution. Use a trigonometric substitution to handle the square root: since $$\sqrt{x$$^{2}$$ - 9}$$ resembles $$\sqrt{x$$^{2}$$ - a$$^{2}$$}$$, set $$x = 3 \sec 	heta$$, which implies $$dx = 3 \sec 	heta 	an 	heta \, d	heta$$ and $$\sqrt{x$$^{2}$$ - 9} = 3 	an 	heta. $$Rewrite the integral in terms of $$	heta$$: substitute $$x$$, $$dx$$, and $$\sqrt{x$$^{2}$$ - 9}$$ into the integral to get $$\int \frac{3 \sec 	heta 	an 	heta \, d	heta}{3 \sec 	heta \cdot 3 	an 	heta}. $$Simplify the expression and then integrate with respect to $$	heta. $$After integration, back-substitute to express the answer in terms of $$t.$$

Evaluate the integrals in Exercises 31–78.
77. ∫dt/((t+1)√(t²+2t-8))

Verified video answer for a similar problem:

Key Concepts

Integration by Substitution

Completing the Square

Standard Integrals Involving Square Roots