Integration — Important Questions

∫1/x dx = log|x| + c, ∫e^x dx = e^x + c, ∫cos x dx = sin x + c, ∫x^(-1) dx = log|x| + c

Standard integration formulas: ∫1/x dx = log|x| + c, ∫e^x dx = e^x + c, ∫cos x dx = sin x + c. The incorrect option is ∫sin x dx = cos x + c; it should be ∫sin x dx = -cos x + c. Also, ∫x^(-1) dx = ∫1/x dx = log|x| + c.

For ∫x√(x² + 1) dx, use u = x² + 1, For ∫sin x cos x dx, use u = sin x, For ∫1/(x log x) dx, use u = log x, For ∫e^(3x) dx, use u = 3x

Good substitutions simplify the integral. u = x² + 1 works because du = 2x dx (x is present). u = sin x works because du = cos x dx (cos x is present). u = log x works because du = dx/x (1/x is present). u = 3x works for e^(3x). For x², no substitution is needed - direct integration suffices.

x²/2 log x - x²/4 + c

Using integration by parts: ∫u dv = uv - ∫v du. Step 1: Let u = log x, dv = x dx. Step 2: du = 1/x dx, v = x²/2. Step 3: ∫x log x dx = (log x)(x²/2) - ∫(x²/2)(1/x) dx. Step 4: = x²/2 log x - ∫x/2 dx = x²/2 log x - x²/4 + c

1/4 log|x-2|/|x+2| + c

Using partial fractions for 1/(x² - 4) = 1/((x-2)(x+2)). Step 1: 1/((x-2)(x+2)) = A/(x-2) + B/(x+2). Step 2: 1 = A(x+2) + B(x-2). Step 3: When x = 2: 1 = 4A, so A = 1/4. When x = -2: 1 = -4B, so B = -1/4. Step 4: ∫[1/4/(x-2) - 1/4/(x+2)] dx = 1/4 log|x-2| - 1/4 log|x+2| + c = 1/4 log|x-2|/|x+2| + c