If _ - ^ 32 2 x^2 (1+7^x ) (x^4+16 ) dx = k , then the value of k is __________

Step-by-step solution

Let I = _ - ^ 32 2 x^2 (1+7^x ) (x^4+16 ) dx We can split the integral as: I = _ - ⁰ 32 2 x^2 (1+7^x ) (x^4+16 ) dx + ₀^ 32 2 x^2 (1+7^x ) (x^4+16 ) dx In the first integral, substitute x = -t dx = -dt . As x - , t , and as x 0, t 0 . _ ⁰ 32 2 (-t)^2 (1+7^ -t ) ((-t)^4+16 ) (-dt) = ₀^ 32 2 t^2 7^t (1+7^t ) (t^4+16 ) dt Replacing t with x , the first integral becomes ₀^ 32 2 x^2 7^x (1+7^x ) (x^4+16 ) dx . Adding this to the second integral: I = ₀^ 32 2 x^2 (7^x + 1 ) (1+7^x ) (x^4+16 ) dx = ₀^ 32 2 x^2 x^4+16 dx Su