Find the antiderivative of x(1 + x^2)^5 ?

Answer 1

antiderivative of [ x(1 + x^2)^5 ]
let u = (1 + x^2)
du/dx = 2x
x = 1/2 du/dx

substitute
x(1 + x^2)^5
[1/2 du/dx] u^ 5

1/2 u^5 du/dx
[ 1/2 u^(5+1) ] / 5+1 +C
(1/2) (1/6) u^6 + C
1/12 u^6 + C

substitute the value of u
1/12 (1 + x^2)^6 + C

Answer 2

Let u=X^2 + 1
(du/dx)=2X

So dx = du/2X

Then putting this in the main problem you will notice that it will become (x(1 + x^2)^5)du/2x. Notice the x at both the denominator and numerator will cancel and you will be left with ((1 + x^2)^5)du/2

but since you had assigned 1+x^2 as U you will have just U^5 du/2 to intergrate.

When you do this you will have (U^6)/12 + K

Now is you put back you stuff the your final answer will be ((1+x^2)^6)/12 + k

.

Answer 3

expand x(1+x^2)^5 =
x(x^10 + 5x^8 + 10x^6+ 10x^4 + 5x^2 + 1)

then the antiderivative is

(1/12)x^12 + (1/2)x^10 + (5/4)x^8 + (5/3)x^6 + (5/4)x^4 + (1/2)x^2 + C

C is a constant

Answer 4

Hint: Make the substitution u = 1 + x^2. Then du = ?