find anti derivative of f(x) 5x^4/3 + 8x^5/4
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
∫ (x2+3) = ∫x2 + ∫3(1/3)X3 + 3X + C
The anti-derivative of 1/x is ln|x| + C, where ln refers to logarithm of x to the base e and |x| refers to the absolute value of x, and C is a constant.
-(1/2)X^2 [negative half X squared]
The anti-derivative of sqrt(x) : sqrt(x)=x^(1/2) The anti-derivative is x^(1/2+1) /(1/2+1) = (2/3) x^(3/2) The anti-derivative is 4e^x is 4 e^x ( I hope you meant e to the power x) The anti-derivative of -sin(x) is cos(x) Adding, the anti-derivative is (2/3) x^(3/2) + 4 e^x + cos(x) + C
The anti-derivative of X2 plus X is the same as the anti-derivative of X2 plus the anti-derivative of X. The anti derivative of X2 is X3/3 plus an integration constant C1 The anti derivative of X is X2/2 plus an integration constant C2 So the anti-derivative of X2+X is (X3/3)+(X2/2)+C1+C2 The constants can be combined and the fraction can combined by using a common denominator leaving (2X3/6)+(3X2/6)+C X2/6 can be factored out leaving (X2/6)(2X+3)+C Hope that helps
find anti derivative of f(x) 5x^4/3 + 8x^5/4
3
x squared
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
∫ (x2+3) = ∫x2 + ∫3(1/3)X3 + 3X + C
The antiderivative of -3 with respect to x is -3x+C. C being any real number.
It is x|x|/2 + C
x (ln x + 1) + Constant
The anti-derivative of 1/x is ln|x| + C, where ln refers to logarithm of x to the base e and |x| refers to the absolute value of x, and C is a constant.
the derivative of 3x is 3 the derivative of x cubed is 3 times x squared