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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{4 u}{( u ^{2} + 3 ) ^{\frac{3}{2}}} d u

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \cdot \int \frac{1}{2 v ^{\frac{3}{2}}} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{3}{2}}} d v

Apply the Power Rule

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{9 x ^{2} - 26 x - 5}{3 x ^{2} - 5 x - 2} d x

sin (7 t + 4)

\int \frac{1}{6 x + 1} d x

\int_{0}^{\frac{1}{2}} \frac{2}{1 + 4 x ^{2}} d x

y^{^{'}} = \frac{2 x ( y - 1 )}{x ^{2} + 3}

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