Question

Make a substitution to express the integrand as a rational function and then evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) x 49 x dx

Answer 1

`\int{(√(x + 49))/(x)} \, dx =2√(x + 49) +49\ln|(√(x + 49)-7)/(√(x + 49)+7)|+c`

Explanation:

Given

`\int\limits {(√(x + 49))/(x)} \, dx`

Required

Solve by substitution

Let:

`u = √(x + 49)`

Square both sides

`u^2 = x + 49`

Differentiate

`2udu = dx`

Also notice that:

`x = u^2 - 49`

So, we have:

`\int\limits {(√(x + 49))/(x)} \, dx =\int\limits {(u)/(u^2 - 49)} \, 2udu`

`\int\limits {(√(x + 49))/(x)} \, dx =\int\limits {(2u^2)/(u^2 - 49)} \, du`

Add 0 to the numerator

`\int\limits {(√(x + 49))/(x)} \, dx =\int\limits {(2u^2+0)/(u^2 - 49)} \, du`

Express 0 as 98 -98

`\int\limits {(√(x + 49))/(x)} \, dx =\int\limits {(2u^2-98+98)/(u^2 - 49)} \, du`

Split the fraction

`\int{(√(x + 49))/(x)} \, dx =\int( (2u^2-98)/(u^2 - 49)+(98)/(u^2 - 49) )\, du`

`\int{(√(x + 49))/(x)} \, dx =\int( (2(u^2-49))/(u^2 - 49)+(98)/(u^2 - 49) )\, du`

`\int{(√(x + 49))/(x)} \, dx =\int( 2+(98)/(u^2 - 49) )\, du`

Rewrite as:

`\int{(√(x + 49))/(x)} \, dx =\int( 2+(98)/(u^2 - 7^2) )\, du`

Integrate

`\int{(√(x + 49))/(x)} \, dx =2u +\int(98)/(u^2 - 7^2) \, du`

Remove constant (98)

`\int{(√(x + 49))/(x)} \, dx =2u +98\int(1)/(u^2 - 7^2) \, du`

As a general rule,

`\int (1)/(x^2 - a^2) \, dx = (1)/(2)\ln|(x-a)/(x+a)|`

So, we have:

`\int (1)/(u^2 - 7^2) \, du = (1)/(2)\ln|(u-7)/(u+7)|`

`\int{(√(x + 49))/(x)} \, dx =2u +98(1)/(u^2 - 7^2) \, du` becomes

`\int{(√(x + 49))/(x)} \, dx =2u +98*(1)/(2)\ln|(u-7)/(u+7)|+c`

`\int{(√(x + 49))/(x)} \, dx =2u +49\ln|(u-7)/(u+7)|+c`

Substitute values for u

`\int{(√(x + 49))/(x)} \, dx =2√(x + 49) +49\ln|(√(x + 49)-7)/(√(x + 49)+7)|+c`