how do i evaluate these integrals?

Question

∫ (x^3/(square root of 16-x^2) ) dx
∫ (1/(square root of x^2+2x+2)) dx
∫ (x/ e^2x)dx
∫ (e^square root of x) dx
∫ ((x-9) / (x^2 + 3x - 10)) dx

thanks!

Answer 1

I'm not going to do all five of these for you, but I'll tell you the appropriate approach for each.

For the first, use inverse trigonometric substitution; use the substitution x = 4 sin θ. (The trigonometric identity sin^2 θ + cos^2 θ = 1 will be useful twice in this problem.) To convert from θ back to x after integrating, you may need to draw a right triangle using the information x = 4 sin θ.

For the second, complete the square under the square root, which will give you ∫(1/(√((x + 1)^2 + 1)) dx. Now use inverse trigonometric substitution, with the substitution x + 1 = tan θ.

For the third, rewrite as ∫ xe^(-2x) dx, then use integration by parts.

For the fourth, use the substitution u = √x to convert the integral to ∫2u e^u du. Now use integration by parts.

For the fifth, use the partial fraction decomposition to write ∫((x - 9) / (x^2 + 3x - 10)) dx = ∫(2/(x + 5))dx - ∫(1/(x - 2))dx and use the Log Rule.

Answer 2

?(3x - 2)^20 dx is almost interior the type u^n du u = 3x - 2, du = 3dx. It probable astounding if I had a three available, so i'm only gonna located on there, in spite of the incontrovertible fact that I definitely have gained to stability it out, so I''ll placed a a million/3 out front. (a million/3)? (3x - 2)^20 3dx now i've got u^du, i will now do u^(n + a million)/(n + a million) (a million/3)(3x - 2)^21 / 21 simplify... (a million/sixty 3)(3x - 2)^21 + C

Answer 3

shane I’m not sure you would understand above 1 star, so I’ll work out only 1 integral; thus y(x)*dx = dx*x^3/√(16-x^2);
♣ x=4*sint, dx=dt*4*cost;
y(t)*dt = dt*cost*(4sint)^3 /√(1-(sint)^2)=
= 64dt*(sint)^3 =64dt*(1-(cost)^2)*sint =
= 64dt*sint -64dt*sint *(cost)^2; hence
♦ Y(t)= -64cost +64dt*(cost)^3 /3 =
Y(x) = -16√(16 -x^2) +(1/3)*√(16 -x^2)^3 +C;
Respond if you get it, so I could continue;