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The integral of f'(x) = 1 is f(x) = x + c

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Using chain rule:integral of cos2x dx= 1/2 * sin2x + C


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The integration formulas covered in the second PUC syllabus primarily include basic integration techniques such as integration of power functions, trigonometric functions, exponential functions, and logarithmic functions. Key formulas include ∫ x^n dx = (x^(n+1))/(n+1) + C for n ≠ -1, ∫ sin(x) dx = -cos(x) + C, and ∫ e^x dx = e^x + C. Additionally, students learn about integration by substitution and integration by parts. Understanding these fundamental formulas is essential for solving various problems in calculus.


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What is the integral of 1 divided by x with respect to x?

∫ (1/x) dx = ln(x) + C C is the constant of integration.


What is the anti-derivative of ln x?

The anti-derivative of ( \ln x ) can be found using integration by parts. Let ( u = \ln x ) and ( dv = dx ), then ( du = \frac{1}{x} dx ) and ( v = x ). Applying integration by parts, we get: [ \int \ln x , dx = x \ln x - \int x \cdot \frac{1}{x} , dx = x \ln x - x + C, ] where ( C ) is the constant of integration. Thus, the anti-derivative of ( \ln x ) is ( x \ln x - x + C ).


What is the indefinite answer of 3(3x-1)4 dx?

To find the indefinite integral of the expression (3(3x-1)^4 , dx), we can use the power rule of integration. First, we can factor out the constant 3, then apply the substitution method or directly integrate. The integral becomes: [ \int 3(3x-1)^4 , dx = 3 \cdot \frac{(3x-1)^5}{15} + C = \frac{(3x-1)^5}{5} + C, ] where (C) is the constant of integration.


What is the integral of x diveded by x minus one?

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What is the integral of tan cubed x secx dx?

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What is the integral of x to the power of n with respect to x?

∫ xn dx = xn+1/(n+1) + C (n ≠ -1) C is the constant of integration.


Integration of cos power of 2x plus sin2x dx?

To integrate the expression ( \cos^2(2x) + \sin(2x) , dx ), we can first rewrite ( \cos^2(2x) ) using the Pythagorean identity: ( \cos^2(2x) = \frac{1 + \cos(4x)}{2} ). The integral then becomes: [ \int \left( \frac{1 + \cos(4x)}{2} + \sin(2x) \right) , dx. ] This simplifies to: [ \frac{1}{2} \int (1 + \cos(4x)) , dx + \int \sin(2x) , dx, ] which can be integrated to yield: [ \frac{1}{2} \left( x + \frac{\sin(4x)}{4} \right) - \frac{1}{2} \cos(2x) + C, ] where ( C ) is the constant of integration.


What is the integral of 1 divided by the cosine squared of x with respect to x?

∫ 1/cos2(x) dx = tan(x) + C C is the constant of integration.