How do you do these two questions?

(a) ⅛ tan⁻¹(¼)

(b) sec x − ln│csc x + cot x│+ C

Step-by-step explanation:

(a) ∫₀¹ x / (16 + x⁴) dx

∫₀¹ (x/16) / (1 + (x⁴/16)) dx

⅛ ∫₀¹ (x/2) / (1 + (x²/4)²) dx

If tan u = x²/4, then sec²u du = x/2 dx

⅛ ∫ sec²u / (1 + tan²u) du

⅛ ∫ du

⅛ u + C

⅛ tan⁻¹(x²/4) + C

Evaluate from x=0 to x=1.

⅛ tan⁻¹(1²/4) − ⅛ tan⁻¹(0²/4)

⅛ tan⁻¹(¼)

(b) ∫ (sec³x / tan x) dx

Multiply by cos x / cos x.

∫ (sec²x / sin x) dx

Pythagorean identity.

∫ ((tan²x + 1) / sin x) dx

Divide.

∫ (tan x sec x + csc x) dx

Split the integral

∫ tan x sec x dx + ∫ csc x dx

Multiply second integral by (csc x + cot x) / (csc x + cot x).

∫ tan x sec x dx + ∫ csc x (csc x + cot x) / (csc x + cot x) dx

Integrate.

sec x − ln│csc x + cot x│+ C

207/8 is the right

i believe that the answer is

b. y-1=-2(x-4)

hope this !