What is the integral of (3+4x)/(sqrt(7+3x+2x^2)) ?

The integral of (3+4x)/(sqrt(7+3x+2x^2)) is 2sqrt(2x^2+3x+7)+C

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integral of (3+4x)/(sqrt(7+3x+2x^2))

\int \frac{3 + 4 x}{7 + 3 x + 2 x ^{2}} d x

2 2 x^{2} + 3 x + 7 + C

Solution steps

\int \frac{3 + 4 x}{7 + 3 x + 2 x ^{2}} d x

Apply u-substitution

: \int \frac{u}{2 u ^{2} + 47} d u

= \int \frac{u}{2 u ^{2} + 47} 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} + 47} 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 ((3 + 4 x)^{2} + 47)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot 2 ((3 + 4 x)^{2} + 47)^{\frac{1}{2}} : 2 2 x^{2} + 3 x + 7

= 2 2 x^{2} + 3 x + 7

Add a constant to the solution

= 2 2 x^{2} + 3 x + 7 + C

x \to - 2 lim (\frac{x + 2}{x ^{2} - 4})

x \to 3 lim (- (3 - x))

f (x) = - 2 x^{2} - 20 x - 48

f (x) = - 4 cos (x), at x = \frac{π}{2}

\int x^{2} e^{2} x d x

What is the integral of (3+4x)/(sqrt(7+3x+2x^2)) ?
The integral of (3+4x)/(sqrt(7+3x+2x^2)) is 2sqrt(2x^2+3x+7)+C