What is the integral of (2x+2)/(sqrt(2x^2+4x+5)) ?

The integral of (2x+2)/(sqrt(2x^2+4x+5)) is sqrt(2x^2+4x+5)+C

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integral of (2x+2)/(sqrt(2x^2+4x+5))

\int \frac{2 x + 2}{2 x ^{2} + 4 x + 5} d x

2 x^{2} + 4 x + 5 + C

Solution steps

\int \frac{2 x + 2}{2 x ^{2} + 4 x + 5} d x

Apply u-substitution

: \int \frac{u}{2 u ^{2} + 6} d u

= \int \frac{u}{2 u ^{2} + 6} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u}{u ^{2} + 6} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= \frac{1}{2} \cdot \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= \frac{1}{2} \cdot \frac{1}{2} \cdot 2 v^{\frac{1}{2}}

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} \cdot 2 ((2 x + 2)^{2} + 6)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot 2 ((2 x + 2)^{2} + 6)^{\frac{1}{2}} : 2 x^{2} + 4 x + 5

= 2 x^{2} + 4 x + 5

Add a constant to the solution

= 2 x^{2} + 4 x + 5 + C

\int 25 - x^{2} d x

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What is the integral of (2x+2)/(sqrt(2x^2+4x+5)) ?
The integral of (2x+2)/(sqrt(2x^2+4x+5)) is sqrt(2x^2+4x+5)+C