What is the integral of (1+4x)/(sqrt(1+3x+6x^2)) ?

The integral of (1+4x)/(sqrt(1+3x+6x^2)) is 1/(3sqrt(2))sqrt(48x^2+24x+8)+C

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integral of (1+4x)/(sqrt(1+3x+6x^2))

\int \frac{1 + 4 x}{1 + 3 x + 6 x ^{2}} d x

\frac{1}{3 2} 48 x^{2} + 24 x + 8 + C

Solution steps

\int \frac{1 + 4 x}{1 + 3 x + 6 x ^{2}} d x

Apply u-substitution

: \int \frac{u}{2 3 u ^{2} + 5} d u

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

Apply u-substitution

: \int \frac{1}{6 v} d v

= \frac{1}{2} \cdot \int \frac{1}{6 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}{6} \cdot \int \frac{1}{v} d v

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{1}{6} \cdot 2 v^{\frac{1}{2}}

Substitute back

= \frac{1}{2} \cdot \frac{1}{6} \cdot 2 (3 (1 + 4 x)^{2} + 5)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{6} \cdot 2 (3 (1 + 4 x)^{2} + 5)^{\frac{1}{2}} : \frac{1}{3 2} 48 x^{2} + 24 x + 8

= \frac{1}{3 2} 48 x^{2} + 24 x + 8

Add a constant to the solution

= \frac{1}{3 2} 48 x^{2} + 24 x + 8 + C

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What is the integral of (1+4x)/(sqrt(1+3x+6x^2)) ?
The integral of (1+4x)/(sqrt(1+3x+6x^2)) is 1/(3sqrt(2))sqrt(48x^2+24x+8)+C