Question

Pls integrate it.....

Answer 1

hey there,
f(x) = y = (x^2 + 1)^(2 - 3x)

Use logarithmic differentiation:

ln y = ln (x^2 + 1)^(2 - 3x) = (2 - 3x) ln (x^2 + 1)
1/y dy/dx = -3 ln (x^2 + 1) + (2 - 3x)/(x^2 + 1) • 2x = -3 ln (x^2 + 1) + 2x(2 - 3x)/(x^2 + 1)

So, dy/dx = y[-3 ln (x^2 + 1) + 2x(2 - 3x)/(x^2 + 1)]

You can also write it with f(x) = (x^2 + 1)^(2 - 3x) replacing the y in the answer.

Hope this helps!

Answer 2

Answer:

Step-by-step explanation:

∫x+1*√x+2/x-2 dx

∫x+1*x+2/√x^2-4

∫x^2+3x+2/√x^2-4

∫x^2+3x/√x^2-4 +2∫1/√x^2-4

∫x^2+3x/√x^2-4  + 2ln x+√x^2-4

∫x^2/√x^2-4 +∫3x/√x^2-4 +2ln x+√x2-4

∫x^2-4+4 / √x^2-4  +3∫x/√x^2-4 +2ln x+ √x^2-4

∫√x^2 -4 dx + 4∫1/√x^2-4 +3∫x/√x^2-4 +2ln x+ √x^2-4

we know that ∫ √x^2-a^2 dx - x√x^2-a^2 /2 -a^2/2 ln x+√x^2-4

therefore ∫√x^2-4 dx = x√x^2-4 /2 -2 ln x+√x^2-a^2

= x√x^2-4 /2 -2 ln x+√x^2-a^2 + 4ln x+√x^2-a^2 +3∫x/√x^2-4 + 2ln x+ √x^2-4

=x√x^2-4 /2 +3/2∫2x/√x^2-4 +4ln x+√x^2-4

=x√x^2-4 / 2 +3/2∫ dt / √t  +4lnx+√x^2-4

=x√x^2-4 /2 +3 ∫dt/2√t +4lnx+√x^2-4

=x√x^2-4 /2 +3√x^2-4 + 4ln x+√x^2-4

=x√x^2-4 /2 + 6√x^2-4 /2 +4ln x+√x^2-4

1/2(x+6)(√x^2-4) + 4ln(x+√x^2-4 )