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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Simplify

\frac{3 + u}{2 + u} : \frac{3 + u}{2 + u}

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

Apply u-substitution

: \int - \frac{2 v ^{2}}{( 1 - v ^{2} ) ^{\frac{3}{2}} 1 - v ^{2}} d v

= \frac{1}{2} \cdot \int - \frac{2 v ^{2}}{( 1 - v ^{2} ) ^{\frac{3}{2}} 1 - v ^{2}} d v

Take the constant out:

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

= \frac{1}{2} (- 2 \cdot \int \frac{v ^{2}}{( 1 - v ^{2} ) ^{\frac{3}{2}} 1 - v ^{2}} d v)

(1 - v^{2})^{\frac{3}{2}} = (1 - v^{2}) 1 - v^{2}

= \frac{1}{2} (- 2 \cdot \int \frac{v ^{2}}{( 1 - v ^{2} ) 1 - v ^{2} 1 - v ^{2}} d v)

(1 - v^{2}) 1 - v^{2} 1 - v^{2} = (1 - v^{2})^{2}

= \frac{1}{2} (- 2 \cdot \int \frac{v ^{2}}{( 1 - v ^{2} ) ^{2}} d v)

Take the partial fraction of

\frac{v ^{2}}{( 1 - v ^{2} ) ^{2}} : - \frac{1}{4 ( v + 1 )} + \frac{1}{4 ( v + 1 ) ^{2}} + \frac{1}{4 ( v - 1 )} + \frac{1}{4 ( v - 1 ) ^{2}}

= \frac{1}{2} (- 2 \cdot \int - \frac{1}{4 ( v + 1 )} + \frac{1}{4 ( v + 1 ) ^{2}} + \frac{1}{4 ( v - 1 )} + \frac{1}{4 ( v - 1 ) ^{2}} d v)

Apply the Sum Rule:

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

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

\int \frac{1}{4 ( v + 1 )} d v = \frac{1}{4} ln ∣ v + 1 ∣

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

\int \frac{1}{4 ( v - 1 )} d v = \frac{1}{4} ln ∣ v - 1 ∣

\int \frac{1}{4 ( v - 1 ) ^{2}} d v = - \frac{1}{4 ( v - 1 )}

= \frac{1}{2} (- 2 (- \frac{1}{4} ln ∣ v + 1 ∣ - \frac{1}{4 ( v + 1 )} + \frac{1}{4} ln ∣ v - 1 ∣ - \frac{1}{4 ( v - 1 )}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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