another integral?

Question

ok well actually i have 2 more integrals that i cant figure out. if i could get some help i would appreciate it.

x^3(e^(x^2))/(x^2 +1)^2

and

ln((x^2)+1)

i really just need to know how to do these not just an answer

Answer 1

Here is the second integral. Integrate by parts.
∫udv = uv - ∫vdu

∫ln(x²+1)dx

u = ln(x²+1)
du = [2x/(x²+1)]dx
dv = dx
v = x

∫ln(x²+1)dx = xln(x²+1) - ∫[2x²/(x²+1)]dx

For the remaining integral use a trig substitution
Let
tanθ = x
sec²θ dθ = dx

∫[2x²/(x²+1)]dx = ∫[2tan²θ/(tan²θ+1)]sec²θdθ
= ∫[2tan²θ/sec²θ]secθ²dθ = 2∫(sec²θ - 1)dθ = 2tanθ - 2θ + C
= 2x - 2arctan(x) + C

So now we have

∫ln(x²+1)dx = xln(x²+1) - ∫[2x²/(x²+1)]dx
= xln(x²+1) - 2x + 2arctan(x) + C

Answer 2

1♠ y’(x)=x^3(e^(x^2))/(x^2 +1)^2; t=x^2, dt=2xdx;
dy =0.5t*dt*expt*(t+1)^2 = (0.5t^3 +t^2 +0.5t)*expt*dt;
the expected solution is:
♣ y(t)=(0.5t^3 +at^2 +bt +c)*expt,
since differentiating is easier find constants a, b, c;
y’ =(0.5t^3 +(a+3*0.5)t^2 +(2a+b)t +(b+c))*expt,
hence a+1.5=1, 2a+b=0.5, b+c=0;
♦ thus y(t) =(0.5t^3 –0.5t^2 +1.5t –1.5)*expt+C;
y(t) =0.5(t-1)(t^2+3) *expt+C;
☼ y(x)= 0.5(x^2-1)(x^4+3) *exp(x^2) +C;

2♠ NorthStar is correct!

Answer 3

try http://integrals.wolfram.com/index.jsp