1- What is the value of the integral ∫(x)/(√(x) + 1) dx. a)...

1- What is the value of the integral $\int \frac{x}{\sqrt{x} + 1} dx$. a) $-\frac{1}{2}ln(\sqrt{x} + x) + C$ b) $2ln(\sqrt{x} + 1) + C$ c) $-2ln(x^2 + \sqrt{x} + 1) + C$ d) $2(\frac{(x+1)^{\frac{3}{2}}}{3} - \sqrt{x+1}) + C$ e) $ln(\sqrt{x}+1)] + C$

Transcript

00:01 Now we have to integrate each of the following with respect to x.

00:03 First is integral of x cos square of x plus x plus 2 whole square into dx.

00:12 We have to integrate this.

00:14 To integrate this expression we will break down into two separate integrals and then solve each one individually.

00:21 Let's suppose this will be i.

00:22 So this will be equal to integral of x into cos square of x into dx plus second integral will be x plus 2 square into dx.

00:36 Let's solve each integral step by step.

00:39 For this first integral we will use integration by parts.

00:44 That means we use integral of uv dv, uv sorry, we will use integration of u dv that is equal to u into integration of dv that means v minus integral sign derivative of first that is du integration of second is nothing but v.

01:06 So we will use this formula.

01:08 Let our u be equal to x.

01:12 Here for this we are taking this as a first function and this is as a second function because we know the l .i .a .t .e.

01:19 Rule according to which algebraic first comes and trigonometric comes the second.

01:27 Means after algebraic we will move to the trigonometric.

01:30 First preference will be given to algebraic.

01:32 So u is x and dv will be equal to cos square of x dx.

01:38 Then differentiate u and integrate dv.

01:40 So we will get here du is equal to dx.

01:47 Then v is equal to integral of cos square of x dx.

01:54 Since dv is this much that integrating both sides we get v equal to integration of this.

02:00 So this is nothing but integral of 1 plus cos of 2x divided by 2 into dx.

02:06 This is equal to 1 by 2 x plus sin of 2x divided by 2 and this 2 is already there here.

02:19 So separating we get here 1 by 2x plus 1 by 4 sin square of x.

02:27 So this is the required v.

02:29 Now we will apply integration by patch formula.

02:32 So integral over x cos square of x dx will be equal to uv minus integral of vdu.

02:42 This will be equal to uv product.

02:45 That is x into x by x into v.

02:51 That means x into v is nothing but equal to this much quantity.

02:59 Substituting here we get 1 by 2x plus 1 by 4 sin square of x minus integral of v.

03:07 Again we will put 1 by 2x plus sin of 2 sorry here is sin 2x not square.

03:17 Sin 2x divided by 2 and here also sin 2x into du.

03:23 Du is nothing but equal to dx here we have found.

03:26 So we have to just evaluate this and we will get the required answer for first part.

03:31 This is equal to x square divided by 2 plus x into sin of 2x divided by 4.

03:38 So just we will multiplying this x inside then minus 1 by 2 integration of x dx is nothing but x square by 2 then minus plus minus 1 by 2 integration of sin is minus cos of 2x and 2 will divide.

03:54 Then this will be the required and plus c1 as a constant of integration.

04:00 Here was sin 2x divided by 4 so here 1 by 4 will come and this will be required.

04:06 Therefore it is equal to x square by 2 plus x sin of 2x divided by 4 minus x square divided by 4 plus cos of 2x divided by 8 plus 7...

Question

1- What is the value of the integral $\int \frac{x}{\sqrt{x} + 1} dx$. a) $-\frac{1}{2}ln(\sqrt{x} + x) + C$ b) $2ln(\sqrt{x} + 1) + C$ c) $-2ln(x^2 + \sqrt{x} + 1) + C$ d) $2(\frac{(x+1)^{\frac{3}{2}}}{3} - \sqrt{x+1}) + C$ e) $ln(\sqrt{x}+1)] + C$

Please give Ace some feedback