Question

Solve the given integral by substituting. ?

Answer 1

`(2x+1)^(3/2)(3x-1)/15+C`

Explanation:

Try `2x+1=u^2`. Then `x = (u^2-1)/2` and `dx = udu`
So, we have

`int x(2x+1)^(1/2)dx = int (u^2-1)/2 times u times u du=1/2 int(u^4-u^2) = 1/10u^5-1/6u^3+C = u^3(u^2/10-1/6)+C = (2x+1)^(3/2)((2x+1)/10-1/6)+C=(2x+1)^(3/2)(3x-1)/15+C`

Answer 2

`1/15(2x+1)^(3/2)(3x-1)+C`.

Explanation:

Let us solve the Problem without using the substitution.

We will use this Rule :

`intf(x)dx=F(x)+c rArr intf(ax+b)dx=1/aF(ax+b)+c', (a!=0).`

E.g., `intx^(1/2)dx=x^(1/2+1)/(1/2+1)=2/3x^(3/2)dx+c_1`,

`rArr int(2x+1)^(1/2)dx=1/2*2/3(2x+1)^(3/2)+c_2`.

`I=intx(2x+1)^(1/2)dx`,

`=1/2int(2x)(2x+1)^(1/2)dx`,

`=1/2int{(2x+1)-1}(2x+1)^(1/2)dx`,

`=1/2int{(2x+1)(2x+1)^(1/2)-(2x+1)^(1/2)}dx`,

`=1/2[int(2x+1)^(3/2)dx-int(2x+1)^(1/2)dx]`,

`=1/2[1/2*2/5(2x+1)^(5/2)-1/2*2/3(2x+1)^(3/2)]`,

`=1/10(2x+1)^(5/2)-1/6(2x+1)^(3/2)`,

`=1/30(2x+1)^(3/2){3(2x+1)-5}`.

`rArr I=1/30(2x+1)^(3/2)(6x-2)=1/15(2x+1)^(3/2)(3x-1)+C`.

Enjoy Maths!