Evaluate the following integral # x^3sqrt(1+x^2)#?
3 Answers
To integrate by
Explanation:
Let
This make
# = 1/2 int (u^(3/2)-u^(1/2)) du#
# = 1/2[2/5u^(5/2) - 2/3 u^(3/2)] +C#
# = 1/15 [3(1+x^2)^(5/2) - 5(1+x^2)^(3/2)] +C#
# = 1/15 (1+x^2)^(3/2)[3+3x^2-5 ] +C#
# = 1/15 (x^2-1)^(3/2) ( 3x^2-2) +C#
To integrate by parts, see below.
Explanation:
We can integrate
#int x^2 (1+x^2)^(1/2) x dx#
Let
so
We can evaluate the integral by substitution, to get
# = 1/3 x^2 (1+x^2)^(3/2) - 2/3 1/5 (1+x^2)^(5/2) +C#
# = 1/15(x^2 +1)^(3/2)[5x^2-2(x^2+1)]#
# = 1/15(x^2 +1)^(3/2)(3x^2-2)#
Explanation:
Let me try another approach:
#d/(dx) x^n (1+x^2)^(3/2) = nx^(n-1) (1+x^2)^(3/2) + 3/2x^n(2x) (1+x^2)^(1/2)#
#color(white)(d/(dx) x^n (1+x^2)^(3/2)) = nx^(n-1) (1+x^2)^(3/2) + 3x^(n+1) (1+x^2)^(1/2)#
#color(white)(d/(dx) x^n (1+x^2)^(3/2)) = (nx^(n-1) (1+x^2)+3x^(n+1))(1+x^2)^(1/2)#
#color(white)(d/(dx) x^n (1+x^2)^(3/2)) = ((n+3)x^(n+1)+nx^(n-1))(1+x^2)^(1/2)#
In particular, putting
#d/(dx) (1/5x^2(1+x^2)^(3/2)) = (x^3+2/5x)(1+x^2)^(1/2)#
#d/(dx) (-2/15(1+x^2)^(3/2)) = -2/5x(1+x^2)^(1/2)#
Hence:
#int x^3(1+x^2)^(1/2) dx = (1/5x^2-2/15)(1+x^2)^(3/2) + C#
#color(white)(int x^3(1+x^2)^(1/2) dx) = 1/15(3x^2-2)(1+x^2)^(3/2) + C#