The exact value of the integral ₀^ x 2 + x dx is:

Step-by-step solution

Let I = ₀^ x 2 + x dx . Using the property ₀^a f(x) dx = ₀^a f(a-x) dx , we get: I = ₀^ - x 2 + ( - x) dx = ₀^ - x 2 + x dx Adding the two expressions for I : 2I = ₀^ x + - x 2 + x dx = ₀^ dx 2 + x I = 2 ₀^ dx 2 + x Now, substitute ( x 2 ) = t . Then dx = 2dt 1+t^2 and x = 2t 1+t^2 . The limits change from x = 0 to t = 0 . I = 2 ₀^ 2dt 1+t^2 2 + 2t 1+t^2 = 2 ₀^ 2dt 2(1+t^2) + 2t I = 2 ₀^ dt t^2 + t + 1 Completing the square in the denominator: I = 2 ₀^ dt (t + 1 2 )^2 + 3 4 Using the standard integral formula dx x^