Integration help (1 Viewer)

[housemouse]

Could you guys start me off with the following integrals (suggest a substitution):

xlog(x^2 + 1) dx / x^2 +1

 

[Raginsheep]

u=log(x²+1)

 

[Riviet]

du=(2x dx )/(x²+1)
Substitute u and du into your original integral:
2∫u du
You can do the rest.

Advice: look for a function and its derivative so you can use a substition just like in this question.

 

[housemouse]

Thanks for that.

Heres some more:

How would you differentiate 2^x?

and

find the integral of x^2 dx /2 +2X^2

 

[Riviet]

d/dx (2^x)=2^x.ln2 (you can prove this by differentiating e^log_e2x)

 

[insert-username]

How would you differentiate 2^x?

I'm not sure if you're required to know it, but the derivative of any exponential function is the function itself times the log_e of its base (remember when you were learning about the exponential function, and you were told that the derivative is equal to a constant times the function. The constant is the log_e of the base), i.e.

d/dx (2^x) = 2^x.log_e2

find the integral of x^2 dx /2 +2X^2

∫(x²dx)/(2 + 2x²)

= 1/2∫(x²dx)/(1 + x²)

Dividing denominator into numerator, we get:

1/2∫(1 - 1/(1+ x²) dx

This can now be integrated:

= x/2 - 1/2tan^-1x + c


I_F

 

[Riviet]

For the integral, factorise the denominator and take the two out leaving you with:
x²/(x²+1) = (x²+1-1)/(x²+1)
= (x²+1)/(x²+1) - 1/(x²+1)
= 1 - 1/(x²+1)
The rest is easy.

 

[SoulSearcher]

I'm not sure if you're required to know it, but the derivative of any exponential function is the function itself times the loge of its base (remember when you were learning about the exponential function, and you were told that the derivative is equal to a constant times the function.

It's still in the syllabus, so theoretically you should know it, even though it isn't tested much.

 

[housemouse]

Integrate sec(x) dx

 

[Trev]

int. sec(x) . dx
Multiply numerator [sec(x)] and denominator [1] by 'sec(x)+tan(x)'.
int. [sec²(x)+sec(x)tan(x)]/[sec(x)+tan(x)] . dx
(I'll show this by substitution, but it is an integration question that you would not know how to integrate without having seen it before, or using a table of standard integrals).
Let u=sec(x)+tan(x)
du/dx = tan(x)sec(x)+sec²(x)
du=[tan(x)sec(x)+sec²(x)].dx
So the integral becomes:
int. 1/u . du
= ln + C
= ln[sec(x)+tan(x)] + C.