∫[t5/√(t2+7)dt Let t = √7tanθ Then dt = √7sec2θdθ
= 49√7∫[tan5θ √7sec2θ/(√7secθ)]dθ
= 49√7∫(sec2θ-1)2tanθsecθdθ Let u = secθ So, du = secθtanθdθ
= 49√7[(1/5)u5 - (2/3)u3 + u] + C
= 49√7[(1/5)sec5θ - (2/3)sec3θ + secθ] + C (**)
To express the answer in terms of the original variable, t, use the fact that t = √7tanθ to get tanθ = t/√7. Then draw a right triangle with acute angle θ, label the side opposite θ as t, the side adjacent to θ as √7, and label the hypotenuse √(t2+7) (by the Pythagorean Theorem). So, secθ = √[(t2+7)/7]. Substitute for secθ in expression (**) and that's your answer.