Easy Problems
$$\int 3x^2 \, dx$$
Solution:
Using the power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C
$$\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} + C = x^3 + C$$
$$\int (4x - 5) \, dx$$
$$\int (4x - 5) \, dx = 2x^2 - 5x + C$$
$$\int e^x \, dx$$
$$\int e^x \, dx = e^x + C$$
Medium Problems
$$\int \sin(3x) \, dx$$
Let u = 3x, du = 3dx
$$\int \sin(3x) \, dx = -\frac{1}{3}\cos(3x) + C$$
$$\int x \cdot e^{x^2} \, dx$$
Let u = x², then du = 2x dx
$$\int x \cdot e^{x^2} \, dx = \frac{1}{2}e^{x^2} + C$$
$$\int_0^1 x^2 \, dx$$
$$\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} - 0 =
\frac{1}{3}$$
Hard Problems
$$\int x \cdot \ln(x) \, dx$$
Integration by parts: u = ln(x), dv = x dx
$$\int x \ln(x) \, dx = \frac{x^2}{2}\ln(x) - \frac{x^2}{4} + C$$
$$\int \frac{1}{x^2 - 1} \, dx$$
Partial fractions: 1/(x²-1) = A/(x-1) + B/(x+1)
$$\int \frac{1}{x^2 - 1} \, dx = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C$$
📚 Textbook Problems (NCERT Exercise 7.4)
Integrate the following functions:
$$\int \frac{3x^2}{x^6 + 1} \, dx$$
Let u = x³, then du = 3x² dx
$$= \int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u) + C = \tan^{-1}(x^3) + C$$
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Step-by-Step Solution →
$$\int \frac{1}{\sqrt{1 + 4x^2}} \, dx$$
Use standard form ∫1/√(a² + x²) dx = sinh⁻¹(x/a) or ln|x + √(x² + a²)|
$$= \frac{1}{2}\sinh^{-1}(2x) + C = \frac{1}{2}\ln|2x + \sqrt{1+4x^2}| + C$$
$$\int \frac{1}{\sqrt{(2-x)^2 + 1}} \, dx$$
Let u = 2 - x, du = -dx
$$= -\sinh^{-1}(2-x) + C = -\ln|(2-x) + \sqrt{(2-x)^2+1}| + C$$
$$\int \frac{1}{\sqrt{9 - 25x^2}} \, dx$$
Rewrite as √(9 - 25x²) = √[9(1 - 25x²/9)] = 3√(1 - (5x/3)²)
$$= \frac{1}{5}\sin^{-1}\left(\frac{5x}{3}\right) + C$$
$$\int \frac{3x}{1 + 2x^4} \, dx$$
Let u = √2·x², then du = 2√2·x dx
$$= \frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}x^2) + C$$
$$\int \frac{x^2}{1 - x^6} \, dx$$
Let u = x³, then du = 3x² dx. Use ∫1/(1-u²)du = (1/2)ln|(1+u)/(1-u)|
$$= \frac{1}{6}\ln\left|\frac{1+x^3}{1-x^3}\right| + C$$
$$\int \frac{x-1}{\sqrt{x^2 - 1}} \, dx$$
Split: ∫x/√(x²-1)dx - ∫1/√(x²-1)dx
$$= \sqrt{x^2-1} - \ln|x + \sqrt{x^2-1}| + C$$
$$\int \frac{x^2}{\sqrt{x^6 + a^6}} \, dx$$
Let u = x³, du = 3x² dx
$$= \frac{1}{3}\ln|x^3 + \sqrt{x^6 + a^6}| + C$$
$$\int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} \, dx$$
Let u = tan x, du = sec²x dx
$$= \ln|\tan x + \sqrt{\tan^2 x + 4}| + C$$
$$\int \frac{1}{\sqrt{x^2 + 2x + 2}} \, dx$$
Complete the square: x² + 2x + 2 = (x+1)² + 1
$$= \ln|(x+1) + \sqrt{x^2+2x+2}| + C$$
$$\int \frac{1}{9x^2 + 6x + 5} \, dx$$
Complete the square: 9x² + 6x + 5 = 9(x + 1/3)² + 4
$$= \frac{1}{6}\tan^{-1}\left(\frac{3x+1}{2}\right) + C$$
$$\int \frac{1}{\sqrt{7 - 6x - x^2}} \, dx$$
Complete the square: 7 - 6x - x² = 16 - (x+3)²
$$= \sin^{-1}\left(\frac{x+3}{4}\right) + C$$
$$\int \frac{1}{\sqrt{(x-1)(x-2)}} \, dx$$
Expand: (x-1)(x-2) = x² - 3x + 2 = (x - 3/2)² - 1/4
$$= \ln\left|x - \frac{3}{2} + \sqrt{(x-1)(x-2)}\right| + C$$
$$\int \frac{1}{\sqrt{8 + 3x - x^2}} \, dx$$
Complete the square: 8 + 3x - x² = 41/4 - (x - 3/2)²
$$= \sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right) + C$$
$$\int \frac{1}{\sqrt{(x-a)(x-b)}} \, dx$$
Let (x-a)(x-b) = (x - (a+b)/2)² - ((a-b)/2)²
$$= \ln\left|x - \frac{a+b}{2} + \sqrt{(x-a)(x-b)}\right| + C$$
$$\int \frac{4x+1}{\sqrt{2x^2 + x - 3}} \, dx$$
Write 4x + 1 = A·(d/dx)(2x² + x - 3) + B = A(4x + 1) + B
$$= 2\sqrt{2x^2+x-3} + C$$
$$\int \frac{x+2}{\sqrt{x^2 - 1}} \, dx$$
Split into two integrals
$$= \sqrt{x^2-1} + 2\ln|x + \sqrt{x^2-1}| + C$$
$$\int \frac{5x-2}{1 + 2x + 3x^2} \, dx$$
Write 5x - 2 = A(6x + 2) + B
$$= \frac{5}{6}\ln|1+2x+3x^2| -
\frac{11}{3\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C$$
$$\int \frac{6x+7}{\sqrt{(x-5)(x-4)}} \, dx$$
Express 6x + 7 in terms of derivative of (x-5)(x-4)
$$= 6\sqrt{(x-5)(x-4)} + 34\ln\left|x-\frac{9}{2}+\sqrt{(x-5)(x-4)}\right| + C$$
$$\int \frac{x+2}{\sqrt{4x - x^2}} \, dx$$
Complete the square: 4x - x² = 4 - (x-2)²
$$= -\sqrt{4x-x^2} + 4\sin^{-1}\left(\frac{x-2}{2}\right) + C$$
$$\int \frac{x+2}{\sqrt{x^2 + 2x + 3}} \, dx$$
Write x + 2 = (1/2)(2x + 2) + 1
$$= \sqrt{x^2+2x+3} + \ln|(x+1) + \sqrt{x^2+2x+3}| + C$$
$$\int \frac{x+3}{x^2 - 2x - 5} \, dx$$
Write x + 3 = (1/2)(2x - 2) + 4
$$= \frac{1}{2}\ln|x^2-2x-5| +
\frac{4}{2\sqrt{6}}\ln\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right| + C$$
$$\int \frac{5x+3}{\sqrt{x^2 + 4x + 10}} \, dx$$
Write 5x + 3 = (5/2)(2x + 4) - 7
$$= 5\sqrt{x^2+4x+10} - 7\ln|(x+2) + \sqrt{x^2+4x+10}| + C$$
Multiple Choice Questions
$$\int \frac{dx}{x^2 + 2x + 2} \text{ equals}$$
(A) x tan⁻¹(x + 1) + C (B) tan⁻¹(x + 1) + C
(C) (x + 1) tan⁻¹x + C (D) tan⁻¹x + C
Complete the square: x² + 2x + 2 = (x+1)² + 1
$$= \tan^{-1}(x+1) + C$$
Answer: (B)
$$\int \frac{dx}{\sqrt{9x - 4x^2}} \text{ equals}$$
(A) (1/9)sin⁻¹((9x-8)/8) + C (B) (1/2)sin⁻¹((8x-9)/9) + C
(C) (1/3)sin⁻¹((9x-8)/8) + C (D) (1/2)sin⁻¹((9x-8)/9) + C
Complete the square: 9x - 4x² = (81/16) - 4(x - 9/8)²
$$= \frac{1}{2}\sin^{-1}\left(\frac{8x-9}{9}\right) + C$$
Answer: (B)