Advertisements
Advertisements
प्रश्न
Find the following integrals:
`int(sqrtx - 1/sqrtx)^2 dx`
उत्तर
Let I = `int (sqrtx - 1/sqrtx)^2 dx`
`I= int (x + 1/x - 2) dx`
`I = int x dx + int 1/x dx - 2 int 1 dx`
`I= x^2/2 + log abs x - 2x + C`
APPEARS IN
संबंधित प्रश्न
Find an anti derivative (or integral) of the following function by the method of inspection.
(axe + b)2
Find the following integrals:
`int (4e^(3x) + 1)`
Find the following integrals:
`int (x^3 + 5x^2 -4)/x^2 dx`
Find the following integrals:
`int (x^3 + 3x + 4)/sqrtx dx`
Find the following integrals:
`int (x^3 - x^2 + x - 1)/(x - 1) dx`
Find the following integrals:
`int(1 - x) sqrtx dx`
Integrate the function:
`1/(xsqrt(ax - x^2)) ["Hint : Put x" = a/t]`
Integrate the function:
`1/(x^2(x^4 + 1)^(3/4))`
Integrate the function:
`sinx/(sin (x - a))`
Integrate the function:
`cos x/sqrt(4 - sin^2 x)`
Integrate the function:
`(sin^8 x - cos^8 x)/(1-2sin^2 x cos^2 x)`
Integrate the function:
`1/(cos (x+a) cos(x+b))`
Integrate the function:
`1/((x^2 + 1)(x^2 + 4))`
Integrate the function:
`cos^3 xe^(log sinx)`
Integrate the function:
f' (ax + b) [f (ax + b)]n
Integrate the function:
`1/sqrt(sin^3 x sin(x + alpha))`
Integrate the function:
`sqrt((1-sqrtx)/(1+sqrtx))`
Integrate the function:
`(2+ sin 2x)/(1+ cos 2x) e^x`
Evaluate `int(x^3+5x^2 + 4x + 1)/x^2 dx`
Evaluate: `int (1 - cos x)/(cos x(1 + cos x)) dx`
The anti derivative of `(sqrt(x) + 1/sqrt(x))` is equals:
`int (dx)/(sin^2x cos^2x) dx` equals
`int (sin^2x - cos^2x)/(sin^2x cos^2x) dx` is equal to
`int (e^x (1 + x))/(cos^2 (xe^x)) dx` equal
`int (dx)/sqrt(9x - 4x^2)` equals
`int (xdx)/((x - 1)(x - 2))` equals
`int (dx)/(x(x^2 + 1))` equals
`int sqrt(1 + x^2) dx` is equal to
What is anti derivative of `e^(2x)`
If the normal to the curve y(x) = `int_0^x(2t^2 - 15t + 10)dt` at a point (a, b) is parallel to the line x + 3y = –5, a > 1, then the value of |a + 6b| is equal to ______.
`int (dx)/sqrt(5x - 6 - x^2)` equals ______.