How do you find the integral of #(7x^(3) -5) (2x^(2) +9) dx#?

Question

Ezra Adams · Accepted Answer

You can multiply the two brackets to get:
#int(14x^5+63x^3-10x^2-45)dx=#
and integrate, separating each term of your integral as a single integral (integral of a sum/difference is equal to the sum/difference of integrals) and using the fact that: #intkx^ndx=kx^(n+1)/(n+1)+c# where #k# is a constant and that #intkdx=kx+c#

Evelyn Agard · Answer

undefined To find the integral of the expression (7x^3 - 5)(2x^2 + 9) dx, you can expand the expression and then integrate each term separately. The integral of a sum is the sum of the integrals. So, first expand the expression using the distributive property:

(7x^3 - 5)(2x^2 + 9) = 14x^5 + 63x^3 - 10x^2 - 45

Then, integrate each term:

∫(14x^5 + 63x^3 - 10x^2 - 45) dx

= ∫14x^5 dx + ∫63x^3 dx - ∫10x^2 dx - ∫45 dx

Now, use the power rule for integration:

∫x^n dx = (1/(n+1)) * x^(n+1) + C

So,

∫14x^5 dx = (14/(5+1)) * x^(5+1) + C = (14/6) * x^6 + C = 7/3 * x^6 + C

∫63x^3 dx = (63/(3+1)) * x^(3+1) + C = (63/4) * x^4 + C

∫10x^2 dx = (10/(2+1)) * x^(2+1) + C = 10/3 * x^3 + C

∫45 dx = 45x + C

Therefore, the integral of (7x^3 - 5)(2x^2 + 9) dx is:

(7/3) * x^6 + (63/4) * x^4 - (10/3) * x^3 - 45x + C