The integral ∫5−4x−x2dx requires converting the expression under the square root into standard form.
The standard formula needed is:
∫a2−u2dx=sin−1(au)+C
To convert 5−4x−x2 into the form a2−u2, complete the square.
Taking out the negative sign from the x2 and x terms:
5−4x−x2=−(x2+4x)+5
Adding and subtracting (24)2=4 inside the bracket:
=−(x2+4x+4−4)+5
=−[(x+2)2−4]+5
=−(x+2)2+4+5
=9−(x+2)2
=32−(x+2)2
The integral becomes:
∫9−(x+2)2dx
This matches the standard form with a=3 and u=(x+2).
Applying the standard formula:
∫9−(x+2)2dx=sin−1(3x+2)+C
Therefore, the answer is sin−1(3x+2)+C