What is the integral of 1/(x^{3/2)+3x+2x^{1/2}} ?

The integral of 1/(x^{3/2)+3x+2x^{1/2}} is -4(1/2 ln|2x^{1/2}+4|-1/2 ln|2x^{1/2}+2|)+C

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integral of 1/(x^{3/2)+3x+2x^{1/2}}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

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

= 2 \cdot \int \frac{1}{( u + \frac{3}{2} ) ^{2} - \frac{1}{4}} d u

Apply u-substitution

: \int \frac{4}{4 v ^{2} - 1} d v

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

Take the constant out:

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

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

Factor

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(x^{3/2)+3x+2x^{1/2}} ?
The integral of 1/(x^{3/2)+3x+2x^{1/2}} is -4(1/2 ln|2x^{1/2}+4|-1/2 ln|2x^{1/2}+2|)+C