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`(1)/(x^(1//2)+x^(1//3))=(1)/(x^(1//3)(1+x^(1//6)))` <br> Let `x=t^(6)impliesdx=6t^(5) dt` <br> ` :. int(1)/(x^(1//2)+x^(1//3))dx=int(1)/(x^(1//3)(1+x^(1//6)))dx`<br> `=int(6t^(5))/(t^(2)(1+t))dt` <br> `=6int(t^(3))/((1+t))dt` <br> On dividing, we obtain <br> `int(1)/(x^(1//2)+x^(1//3))dx=6int{(t^(2)-t+1)-(1)/(1+t)}dt` <br> `=6[((t^(3))/(3))-((t^(2))/(3))+t-log|1+t|]+C` <br> `=2x^(1//2)-3x^(1//3)+6x^(1//6)-6 log(1+x^(1//6))+C`Transcript

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00:00 - 00:59 | hello friends welcome to doctor today we will solve the question evaluate integration of 1 by X raise to power 1 by 2 + X is powered by 3DX it is important to note that we have the term x-powered-by to open my integration of 1 upon X raise to power 1 by 2 + access to powered by 3DX show a weekend at let X is equals to cos x dx is equal to 16 5dt so integration vs 655 DT upon the x is equal to 3 x x raise to power 1 by 2 is equal to 3 days Kochi similarly it is reduced pao2 so this |

01:00 - 01:59 | we can take comment to can be taken out comment so we have tourist 3.3 + 1 into DTH operator and denominator in terms of power we are taking it at so we have to divide the visit so we have to take me take 6 plus one square + b square minus minus out so minus b square minus a minus b square minus plus plus cancelled out |

02:00 - 02:59 | antique antique plus one so plus one minus minus this entry gate considered so -1 so we had the value integration t square - 3 + 1 and -1 by 3 + 1 whole is with it and we have a constant here as 6 the integration of p square is equal to 3 by 3 minus A square by 2 minus 1 by 1 by 3 + 1 integration is log plus one way we can put the value of |

03:00 - 03:59 | we can't put the value as X so here we get the value as 6 into X raise to power 1 by 3 minus x raise to power 1 by 3 by 2 + 1 by 6 - lock X raise to the power 1 by X + 1 + c = 2 X raise to power 1 by 2 minus 3 x power 1 by x axis power 1 by 6 - 6 log x raise to power 1 by 6 + 1 + c minus thank you |

**A function `phi(x)` is called a primitive of `f(x)`; if `phi'(x) = f(x)`**

**Some important formulas of integration**

**Examples of integration: (i) `x^4` (ii) `3^x`**

**Theorem: `d/dx(int f(x) dx) = f(x)`**

**The integral of the product of a constant and a function = the constant x integral of function**

**`int {f(x) pm g(x)} dx = int f(x) dx pm int g(x) dx`**

**Geometrical interpretation of indefinite integral**

**Comparison between differentiation and integration**

**By substitution: Theorem: If `int f(x) dx = phi(x)` then `int f(ax+b) dx = 1/a phi(ax + b)dx`**

**Examples: `1/ (cos3x+1) dx` and `1/((sqrt (x+a) + sqrt (x+b))) dx`**