Integrals Of Special Functions - JEE Maths Free MCQ Test with solutions

∫ [1 / (4x² - 12x + 25)] dx

First, factor out the coefficient of x² from the first two terms:

4x² - 12x + 25 = 4(x² - 3x) + 25

Next, complete the square for the expression inside the parentheses:

x² - 3x = (x - 1.5)² - 2.25

Substitute this back into the original expression:

4[(x - 1.5)² - 2.25] + 25 = 4(x - 1.5)² - 9 + 25 = 4(x - 1.5)² + 16

The integral now becomes:

∫ [1 / (4(x - 1.5)² + 16)] dx = (1/4) ∫ [1 / ((x - 1.5)² + 4)] dx
Recall the standard integral formula:

∫ [1 / (u² + a²)] du = (1/a) tan⁻¹(u/a) + C

In our case:

• u = x - 1.5
• a = 2

Applying the formula:

(1/4) * (1/2) tan⁻¹((x - 1.5)/2) + C = (1/8) tan⁻¹((2x - 3)/4) + C

The integral is:

∫ [1 / (4x² - 12x + 25)] dx = (1/8) tan⁻¹((2x - 3)/4) + C

= tanx + cotx + C