+42+13dz is equal toA) log (a2 +4x +13) +CC) log (2x +4) +CJ/름tan-1(

1.	+42+13dz is equal toA) log (a2 +4x +13) +CC) log (2x +4) +CJ/름tan-1(부)2+4(+4z+13)D)
Answer» x^2+ 4x + 13 = x² + 4x + 4 + 9 = (x + 2)² + 9 Now ∫ dx/[(x + 2)² + 9] Now that should sort of look familiar, it looks like the derivative of arctan(x) which is 1/(x² + 1) Therefore, Let 3u = x + 2 Then we have that: 3du = dx Now your integral becomes: ∫ 3du/[(3u)² + 9]= ∫ 3du/(9u² + 9)= ∫ du/[3(u² + 1)]= (1/3) ∫ du/(u² + 1)= 1/3 arctan(u) But u = (x + 2)/3 Therefore, your answer is: ∫ dx/(x² + 4x + 13) = (1/3)arctan[(x + 2)/3]

+42+13dz is equal toA) log (a2 +4x +13) +CC) log (2x +4) +CJ/름tan-1(부)2+4(+4z+13)D)

Answer»

x^2+ 4x + 13 = x² + 4x + 4 + 9 = (x + 2)² + 9

Now ∫ dx/[(x + 2)² + 9]

Now that should sort of look familiar, it looks like the derivative of arctan(x) which is 1/(x² + 1)

Therefore,

Let 3u = x + 2

Then we have that:

3du = dx

Now your integral becomes:

∫ 3du/[(3u)² + 9]= ∫ 3du/(9u² + 9)= ∫ du/[3(u² + 1)]= (1/3) ∫ du/(u² + 1)= 1/3 arctan(u)

But u = (x + 2)/3

Therefore, your answer is:

∫ dx/(x² + 4x + 13) = (1/3)arctan[(x + 2)/3]