What is the integral of (2+8x)/(sqrt(7+2x+4x^2)) ?

The integral of (2+8x)/(sqrt(7+2x+4x^2)) is 2sqrt(4x^2+2x+7)+C

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integral of (2+8x)/(sqrt(7+2x+4x^2))

\int \frac{2 + 8 x}{7 + 2 x + 4 x ^{2}} d x

2 4 x^{2} + 2 x + 7 + C

Solution steps

\int \frac{2 + 8 x}{7 + 2 x + 4 x ^{2}} d x

Apply u-substitution

: \int \frac{u}{2 u ^{2} + 108} d u

= \int \frac{u}{2 u ^{2} + 108} 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} + 108} 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 + 8 x)^{2} + 108)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot 2 ((2 + 8 x)^{2} + 108)^{\frac{1}{2}} : 2 4 x^{2} + 2 x + 7

= 2 4 x^{2} + 2 x + 7

Add a constant to the solution

= 2 4 x^{2} + 2 x + 7 + C

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What is the integral of (2+8x)/(sqrt(7+2x+4x^2)) ?
The integral of (2+8x)/(sqrt(7+2x+4x^2)) is 2sqrt(4x^2+2x+7)+C