Integral Calculus Tutor
ఇంటిగ్రల్ కాలిక్యులస్ ట్యూటర్
Learn to solve complex integrals step-by-step.
∫ px+qax²+bx+c dx
∫ px+qax²+bx+c dx
Working Rule
పని చేసే పద్ధతి
To solve this, we express the numerator in the form:
px + q = A * ddx(ax² + bx + c) + B. Then, we find the values of A and B, split the integral into two parts, and solve each part separately.
Solved Example: ∫ x+1x²+3x+12 dx
పరిష్కరించిన ఉదాహరణ: ∫ x+1x²+3x+12 dx
Step 1:దశ 1: Let x+1 = A * ddx(x²+3x+12) + B = A(2x+3) + B.
Step 2:దశ 2: Comparing coefficients, 1=2A ⇒ A=12 and 1=3A+B ⇒ B=-12.
Step 3:దశ 3: Split the integral: ∫ 12(2x+3)x²+3x+12 dx - ∫ 12x²+3x+12 dx.
Step 4:దశ 4: The first part is 12log|x²+3x+12|. For the second part, complete the square: x²+3x+12 = (x+32)² + 394.
Step 5:దశ 5: The second integral becomes -12 * 2√39tan⁻¹(x+32√392).
Final Answer:తుది సమాధానం: 12log|x²+3x+12| - 1√39tan⁻¹(2x+3√39) + C.
Try It Yourself!
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Enter the coefficients for ∫ px+qax²+bx+c dx
Type 2: ∫ (px+q)√(ax²+bx+c) dx
రకం 2: ∫ (px+q)√(ax²+bx+c) dx
Working Rule
పని చేసే పద్ధతి
The method is the same as Type 1. Express the linear part in the form:
px + q = A * ddx(ax² + bx + c) + B. This splits the integral into two manageable parts.
Solved Example: ∫ (3x-2)√(2x²-x+1) dx
పరిష్కరించిన ఉదాహరణ: ∫ (3x-2)√(2x²-x+1) dx
Step 1:దశ 1: Let 3x-2 = A(4x-1) + B.
Step 2:దశ 2: Comparing coefficients, A=34 and B=-54.
Step 3:దశ 3: Split: 34∫(4x-1)√(2x²-x+1) dx - 54∫√(2x²-x+1) dx.
Step 4:దశ 4: First part (by substitution t=2x²-x+1) is 34 * 23(2x²-x+1)32.
Step 5:దశ 5: Second part requires completing the square and using the formula for ∫√(x²+a²)dx.
Final Answer:తుది సమాధానం: 12(2x²-x+1)32 - 5(4x-1)32√(2x²-x+1) - ... (and a log term).
Try It Yourself!
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Enter the coefficients for `∫ (px+q)√(ax²+bx+c) dx`
∫ dx(ax+b)√(px+q)
∫ dx(ax+b)√(px+q)
Working Rule
పని చేసే పద్ధతి
For this type, the key is substitution. Let t = √(px + q). Then, square both sides to get t² = px + q. From this, find expressions for `x` and `dx` in terms of `t` and substitute them back into the integral.
Solved Example: ∫ dx(x+5)√(x+4)
పరిష్కరించిన ఉదాహరణ: ∫ dx(x+5)√(x+4)
Step 1:దశ 1: Let t = √(x+4). Then t² = x+4.
Step 2:దశ 2: From this, x = t²-4 and differentiating gives dx = 2t dt.
Step 3:దశ 3: Substitute into the integral: ∫ 2t dt((t²-4)+5) * t.
Step 4:దశ 4: Simplify: ∫ 2t dt(t²+1) * t = ∫ 2t²+1 dt.
Step 5:దశ 5: Integrate to get 2 * tan⁻¹(t) + C.
Final Answer:తుది సమాధానం: 2tan⁻¹(√(x+4)) + C.
Try It Yourself!
మీరే ప్రయత్నించండి!
Enter the coefficients for ∫ dx(ax+b)√(px+q)