2 integration qns?

Question

1. Use the substitution u = e^x +2 to find ∫ e^2x / (e^x +2) dx

2. Find ∫ x^2 / (1 + x^6) dx using the substitution y = x^3

can anyone help me by providing step by step solutions pls.
im stuck and i need help!

Answer 1

1♠ So u=exp(x)+2, → exp(x)=u-2, → x=ln(u-2), dx= du/(u-2);
♣ y*dx= dx*exp(2x) /(exp(x)+2)=
= [du/(u-2)] *(u-2)^2 /u = du*(1-2/u), hence
♦ Y(x) = u -2*ln(u) = exp(x)+2 –2*ln(exp(x)+2) +C;
2♠ so y=x^3, x=y^(1/3), dx= (dy/3)*y^(-2/3); thus
I= (1/3)*∫ dy* y^(-2/3) * y^(2/3) /(1+y^2)=
= (1/3)*∫ dy/(1+y^2)= (1/3)*atan(y)= (1/3)*atan(x^3) +C;

Answer 2

so du=e^2x dx
so I1=∫ udu=u^2/2=((e^x +2)^2)/2

dy=3x^2dx
I2=∫y/3(1+y^2)dy
y^2=z
2ydy=dz

so I2=∫1/6(1+z)dz= 1/6 ln (1+z) = 1/6 ln (1+y^2) =1/6 ln (1+x^6)

Answer 3

Question 1
I am not sure if integral is ∫e^(2x) / e^(x + 2).dx or ∫ e^(2x) / (e^x + 2) dx.
Will go with the former:-
I = ∫e^(2x) / e^(x + 2).dx
I = (1/e²).∫e^2x / e^x dx
I = (1/e²).∫e^x dx
I = (1/e²).e^(x) + C

Question 2
y = x³
dy = 3x² dx
dy/3 = x²dx
I = (1/3).∫1 / (1 + y²) .dy
I = (1/3).tan^(-1) y + C
I = (1/3).tan^(-1) (x³) + C