Question: ∫(x^2)/sqrt(x^2+1)
u=x^2+1 , x^2= u-1, du=2xdx
∫(u-1)/sqrt(u) , expand ∫u/sqrt(u) - 1/sqrt(u)
Integrate: 2/3(u^(3/2)) - 2u^(1/2) + c
My answer: [ 2/3(x^2+1)^(3/2) - 2(x^2+1) + c ]
When I took the derivative of this to check my answer, it was not (x^2)/sqrt(x^2+1). Please help :/
The problem is that you did not have du in your integrals. You lacked the extra 2x dx in your integrands.
u = x^2+1
du = 2x dx
x = √(u-1)
So you have
∫x/√(x^2+1) x dx
= 1/2 ∫√(u-1)/u du
which isn't really much better
Try using a trig substitution:
x = tanθ
x^2+1 = tan^2θ + 1 = sec^2θ
dx = sec^2θ dθ
∫x^2/√(x^2+1) dx
= ∫(sec^2θ-1)/secθ sec^2θ dθ
= ∫sec^3θ-secθ dθ
= 1/2 secθ tanθ - 1/2 log tan(π/4 - θ/2)
= 1/2 x√(x^2+1) - 1/2 log (√(1+x^2) - x)
= 1/2 (x√(x^2+1) - sinh-1x)
To solve this integral, we can start by substituting u = x^2 + 1, which allows us to rewrite the integral as
∫ (x^2)/sqrt(x^2+1) dx = ∫ (u-1)/sqrt(u) dx.
Next, we need to find dx in terms of du to continue the integration. To do this, we can differentiate both sides of the equation u = x^2 + 1 with respect to x:
du/dx = d/dx (x^2 + 1),
du/dx = 2x,
dx = du / (2x).
Now, replacing dx in the integral with du / (2x), we have:
∫ (u-1)/sqrt(u) dx = ∫ (u-1)/sqrt(u) (du / (2x)).
The next step is to rewrite the integral in terms of u only. We separate the fraction into two parts:
∫ (u-1)/sqrt(u) (du / (2x)) = ∫ u/sqrt(u) (du / (2x)) - ∫ 1/sqrt(u) (du / (2x)).
Now, let's simplify each part separately:
∫ u/sqrt(u) (du / (2x)) = (1/2) ∫ (u/u^(1/2)) du = (1/2) ∫ u^(1/2) du = (1/2) * (2/3) u^(3/2) + C1,
∫ 1/sqrt(u) (du / (2x)) = (1/2) ∫ 1/u^(1/2) du = (1/2) * 2 u^(1/2) + C2.
Combining these two integrals, we get:
(1/2) * (2/3) u^(3/2) + C1 - u^(1/2) + C2.
Now, substitute u back in terms of x^2 + 1:
(1/3) (x^2 + 1)^(3/2) + C1 - (x^2 + 1)^(1/2) + C2.
To simplify further, combine C1 and C2 into a single constant C:
= (1/3) (x^2 + 1)^(3/2) - (x^2 + 1)^(1/2) + C.
Thus, my answer to the integral is:
∫ (x^2)/sqrt(x^2+1) dx = (1/3) (x^2 + 1)^(3/2) - (x^2 + 1)^(1/2) + C.
Please check your calculations again and make sure you haven't made any errors in differentiating or integrating.