Use the substitution u=x^2-2 to find the integral of (6x^3+4x)/sqrt( x^2-2)
First use the substitution to find du/dx which is 2x. From this we now know that dx= du/2x (just re-arranging.) Substituting that into the integral we now get (6x3+4x)/ (sqrt (u) x 2x) du. Cancelling out the 2x we now have (3x2+2)/ sqrt (u) du. This is equal to (3u+8)/ sqrt(u) du using the original definition of u. This is equal to 3u0.5+ 8u-0.5 du. Integrating this we get 2u1.5 + 16u0.5+ c and to get the final answer we just substitute for u.
KS
Related Maths A Level answers
All answers ▸find the definite integral between limits 1 and 2 of (4x^3+1)/(x^4+x) with respect to x
y=4x^3+6x+3 so find dy/dx and d^2y/dx^2
How do you differentiate a function comprised of two functions multiplied together?
A curve has equation 3x^4/3-16y^3/4=32. By differentiating implicitly find dy/dx in terms of x and y. Hence find the gradient of the curve at the point (8,1).