What is the integral of (3x)/(sqrt(x^2+2x+5)) ?

The integral of (3x)/(sqrt(x^2+2x+5)) is 3(sqrt(x^2+2x+5)-ln(1/2 |x+1+sqrt(x^2+2x+5)|))+C

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integral of (3x)/(sqrt(x^2+2x+5))

\int \frac{3 x}{x ^{2} + 2 x + 5} d x

3 (x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)) + C

Solution steps

\int \frac{3 x}{x ^{2} + 2 x + 5} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x}{x ^{2} + 2 x + 5} d x

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

= 3 \cdot \int \frac{x}{( x + 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{2} + 4} d u

= 3 \cdot \int \frac{u - 1}{u ^{2} + 4} d u

Expand

\frac{u - 1}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{1}{u ^{2} + 4}

= 3 \cdot \int \frac{u}{u ^{2} + 4} - \frac{1}{u ^{2} + 4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int \frac{u}{u ^{2} + 4} d u - \int \frac{1}{u ^{2} + 4} d u)

\int \frac{u}{u ^{2} + 4} d u = u^{2} + 4

\int \frac{1}{u ^{2} + 4} d u = ln (\frac{1}{2} u + 4 + u^{2})

= 3 (u^{2} + 4 - ln (\frac{1}{2} u + 4 + u^{2}))

Substitute back

u = x + 1

= 3 ((x + 1)^{2} + 4 - ln (\frac{1}{2} x + 1 + 4 + (x + 1)^{2}))

Simplify

3 ((x + 1)^{2} + 4 - ln (\frac{1}{2} x + 1 + 4 + (x + 1)^{2})) : 3 (x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5))

= 3 (x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5))

Add a constant to the solution

= 3 (x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)) + C

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\int \frac{4400 q}{4 q ^{2} + 320} d q

What is the integral of (3x)/(sqrt(x^2+2x+5)) ?
The integral of (3x)/(sqrt(x^2+2x+5)) is 3(sqrt(x^2+2x+5)-ln(1/2 |x+1+sqrt(x^2+2x+5)|))+C