Step 1:Replace any obvious in the integral with the substitution u=x+2 ∫_(-1)^1 3/√u dx

Step 2: Differentiate the substitution u=x+2 to find an expression for dx (to differentiate multiply by the power and then minus one from the power): du/dx=1 du=dx

Step 3: Change the limits and replace in the integral: When x=1, u=1+2=3 When x=-1, u=-1+2=1 So the integral with respect to u looks like this, ∫_(1)^3 3/√u du

Step 4: Rewrite the equation so that it can be integrated, so 3/√u can be written as 3u^-1/2

Step 5: Integrate ∫_(1)^3 3u^-1/2 du (to integrate add one to the power and then divide by the new power, then substitute in the limits subtracting the top from the bottom)and write in its simplest form : ∫_(1)^3 3u^-1/2 du = [6u^1/2]_1^3 = (6x3^1/2)-(6x1^1/2) = 6√3-6√1 = 6√3-6 = 6(√3-1) (final answer)

Factorise the following expression: x^2-1

Related Maths GCSE answers

Solve the simultaneous equations: 4x+5y=13 and 3x-2y=27

Find the value of x which satisfies the following equation 3 x2 + 6 x + 3 = 0

How do you factorise?

Integrate ∫_(-1)^1 3/√(x+2) dx using the substitution u x+2