# Calc.

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Evaluate the indefinite integral. Please check my work? Not sure if I am doing this correctly.

S= integral symbol
S x^3*sqrt(x^2 + 1) dx

u = x^2+1
x^2= u - 1
du = 2xdx
du/2 = xdx
1/2 S x^2 * x sqrt (x^2 + 1) du
1/2 S (u - 1) * sqrt (u) du multiply square root of u and (u - 1)

FOIL (u - 1) * sqrt (u)
=((u)^3/2) - ((u)^1/2)
= 1/2 S (((u)^3/2) - ((u)^1/2)) du Now take the derivative and keep the 1/2 on the side.
1/2 ((u)^5/2)/(5/2) - ((u)^3/2)/(3/2) now distribute the 1/2 and bring the 5/2 and 3/2 to the top
1/2 * 2/5 * (u)^5/2 - 1/2 * 2/3 * (u)^ 3/2
cut all the 2 and substitute x^2+1 in the place of u.
1/5 (x^2 - 1)^5/2 - 1/3(x^2 - 1)^3/2

• Calc. -

| x^3*sqrt(x^2 + 1) dx
| = integral sign

| x^2 (sqrt(x^2 + 1)) x dx

u = x^2
du = 2x dx
1/2 du = x dx

1/2 | u (sqrt(u + 1)) du

w = u + 1
dw = du
u = w - 1

1/2 | (w - 1) (sqrt(w)) dw
1/2 | (w - 1) w^1/2 dw
1/2 | w^3/2 dw - 1/2 | w^1/2 dw

1/2 (2/5 w^5/2) - 1/2 ( 2/3 w^3/2) + C
2/10 w^5/2 - 2/6 w^3/2 + C
1/5 w^5/2 - 1/3 w^3/2 + C

w = u + 1
1/5 (u + 1)^5/2 - 1/3 (u + 1)^3/2 + C

u = x^2
1/5 (x^2 + 1)^5/2 - 1/3 (x^2 + 1)^3/2 +
C