How do I integrate (dx/ (x^2+4) (sqrt (4x^2+1)))?
This is a standard question of the form 1/ quadratic multiplied by square root quadratic.The substitution is x = 1/t and after that one more substitution is required , the quantity in square root must be substituted as u ^2.And the question gets solved.
Integral of (x+3)/(sqrt(5-4x-x^2) dx?
5 - 4x - x^2 => 5 - (x^2 + 4x) => 5 - (x^2 + 4x + 4 - 4) => 5 - (-4) - (x^2 + 4x + 4) => 5 + 4 - (x + 2)^2 => 9 - (x + 2)^2 (x + 3) * dx / sqrt(9 - (x + 2)^2) x + 2 = u dx = du (u + 1) * du / sqrt(9 - u^2) u * du / sqrt(9 - u^2) + du / sqrt(9 - u^2) a = 9 - u^2 da = -2u * du (-1/2) * da / sqrt(a) + du / sqrt(9 - u^2) u = 3 * sin(t) du = 3 * cos(t) * dt (-1/2) * da / sqrt(a) + 3 * cos(t) * dt / sqrt(9 - 9sin(t)^2) => 3 * cos(t) * dt / sqrt(9cos(t)^2) - (1/2) * a^(-1/2) * da => 3 * cos(t) * dt / (3 * cos(t)) - (1/2) * a^(-1/2) * da => dt - (1/2) * a^(-1/2) * da Integrate t - (1/2) * (1/(-1/2 + 1)) * a^(-1/2 + 1) + C => t - (1/2) * (1/(1/2)) * a^(1/2) + C => t - (1/2) * 2 * a^(1/2) + C => t - sqrt(a) + C => arcsin(u/3) - sqrt(9 - u^2) + C => arcsin((x + 2) / 3) - sqrt(5 - 4x - x^2) + C
Integral 1/sqrt(4x-x^2)?
integral 1/sqrt(4 x-x^2) dx For the integrand 1/sqrt(4 x-x^2), complete the square: = integral 1/sqrt(4-(x-2)^2) dx For the integrand 1/sqrt(4-(x-2)^2), substitute u = x-2 and du = dx: = integral 1/sqrt(4-u^2) du The integral of 1/sqrt(4-u^2) is sin^(-1)(u/2): = sin^(-1)(u/2)+constant Substitute back for u = x-2: = sin^(-1)((x-2)/2)+constant Which is equivalent for restricted x values to: = (2 sqrt(x-4) sqrt(x) log(2 (sqrt(x-4)+sqrt(x))))/sqrt(-(x-4) x)+constant
How do you integrate of [math]\frac{2x.dx}{\sqrt{4x-x^2}}[/math]?
∫ 2x/√(4x - x²) dxu = 4x - x²du = (4 - 2x) dx= ∫ (4 - (4 - 2x))/√(4x - x²) dx= 4 ∫ 1/√(4x - x²) dx - ∫ (4 - 2x)/√(4x - x²) dx= (4/i) ∫ 1/√(x² - 4x) dx - ∫ (1/√u) du= (4/i) ∫ 1/√((x² - 4x + 4) - 4) dx - ∫ u^(-1/2) du= (4/i) ∫ 1/√((x - 2)² - 4) dx - 2u^(1/2) + C= (4/i) ∫ 1/√((x - 2)² - 4) dx - 2√u + C= (4/i) ∫ 1/√((x - 2)² - 4) dx - 2√(4x - x²) + Cx - 2 = 2 sec v(x - 2)² = 4 sec²vsec v = (x - 2)/2tan v = (√(x² - 4x))/2dx = 2 sec v tan v dv= (8/i) ∫ 1/√(4 sec²v - 4) sec v tan v dv - 2√(4x - x²) + C= (4/i) ∫ 1/√(sec²v - 1) sec v tan v dv - 2√(4x - x²) + C= (4/i) ∫ 1/(tan v) sec v tan v dv - 2√(4x - x²) + C= (4/i) ∫ sec v dv - 2√(4x - x²) + C= (4/i) ln|sec v + tan v| - 2√(4x - x²) + C= (4/i) ln|((x - 2)/2) + ((√(x² - 4x))/2)| - 2√(4x - x²) + C
How do I integrate (3x-2)(x^2+2x+3^1/2)dx?
integrate((3*x-2)*(x^2+2*x+sqrt(3)),x) in xmaxima gives, upon multthru,3*x^4/4+4*x^3/3+(3*sqrt(3)-4)*x^2/2–2*sqrt(3)*x+k1. You expand out the terms in the integrand and integrate term by term.
How do you integrate the integral sqrt(5+4x-x^2) dx?
∫ √(5 + 4x - x²) dx = complete the square by adding and subtracting 4: ∫ √(5 + 4 - 4 + 4x - x²) dx = ∫ √[(5 + 4) - (x² - 4x + 4)] dx = ∫ √[9 - (x - 2)²] dx = let (x - 2) = t → d(x - 2) = dt → dx = dt substituting, you have: ∫ √(9 - t²) dt = let: dt = dv → t = v √(9 - t²) = u → (1/2)(- 2t)(9 - t²)^[(1/2) -1] dt = (- t)(9 - t²)^(-1/2) dt = [- t /√(9 - t²)] dt = du integrating by parts, you get: ∫ u dv = v u - ∫ v du ∫ √(9 - t²) dt = t √(9 - t²) - ∫ t [- t /√(9 - t²)] dt ∫ √(9 - t²) dt = t √(9 - t²) - ∫ [- t² /√(9 - t²)] dt add and subtract 9 on the top: ∫ √(9 - t²) dt = t √(9 - t²) - ∫ {[(9 - t²) - 9] /√(9 - t²)} dt split it into: ∫ √(9 - t²) dt = t √(9 - t²) - ∫ [(9 - t²)/√(9 - t²)] dt + 9 ∫ [1/√(9 - t²)] dt that simplifies into: ∫ √(9 - t²) dt = t √(9 - t²) - ∫ √(9 - t²) dt + 9 ∫ [1/√(9 - t²)] dt collect ∫ √(9 - t²) dt at the right side: ∫ √(9 - t²) dt + ∫ √(9 - t²) dt = t √(9 - t²) + 9 ∫ [1/√(9 - t²)] dt 2 ∫ √(9 - t²) dt = t √(9 - t²) + 9 ∫ dt /√(9 - t²) ∫ √(9 - t²) dt = (1/2) {t √(9 - t²) + 9 ∫ dt /√(9 - t²)} = (1/2)t √(9 - t²) + (9/2) ∫ dt /√(9 - t²) = factor out 9: (1/2)t √(9 - t²) + (9/2) ∫ dt /√{9[1 - (t²/9)]} = (1/2)t √(9 - t²) + (9/2) ∫ dt /{3 √[1 - (t²/9)]} = rearrange it as: (1/2)t √(9 - t²) + (9/2) ∫ (1/3)dt /√[1 - (t/3)²] = (1/2)t √(9 - t²) + (9/2) ∫ d(t/3) /√[1 - (t/3)²] = thus, recalling that ∫ d[f(x)] /√{1 - [f(x)]²} = arcsin[f(x)] + C, you get: (1/2)t √(9 - t²) + (9/2)arcsin(t/3) + C finally substitute back (x - 2) for u: (1/2)(x - 2)√[9 - (x - 2)²] + (9/2)arcsin[(x - 2)/3] + C = (1/2)(x - 2)√[9 - (x² - 4x + 4)] + (9/2)arcsin[(x - 2)/3] + C = (1/2)(x - 2)√(9 - x² + 4x - 4) + (9/2)arcsin[(x - 2)/3] + C in conclusion: ∫ √(5 + 4x - x²) dx = (1/2)(x - 2)√(5 + 4x - x²) + (9/2)arcsin[(x - 2)/3] + C (I've just checked this successfully on my PC) I hope this helps..
Evaluate the integral *int* x^2sqrt(4x+3)dx?
x^2sqrt(4x+3) dx let u^2 = 4x+3; u = sqrt(4x+3) 2u du = 4 dx 1/2 u du = dx ∫((u^2 - 3)/4)^2 sqrt(u^2) 1/2 u du 1/32 ∫(u^2 - 3)^2 u^2 du 1/32 ∫(u^4 - 6u^2 +9) u^2 du 1/32 ∫u^6 - 6u^4 +9u^2 du 1/32 (1/7) u^7 - (1/32)6(1/5)u^5 + (1/32)9(1/3)u^3 + c 1/224 u^7 - 6/160 u^5 +3/32 u^3 + c 1/224 (4x+3)^7/2 - 3/80 (4x+3)^5/2 +3/32 (4x+3)^3/2 + c