What is the integral of 6x(x^2+1)^{1/2} ?

The integral of 6x(x^2+1)^{1/2} is 2(x^2+1)^{3/2}+C

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integral of 6x(x^2+1)^{1/2}

\int 6 x (x^{2} + 1)^{\frac{1}{2}} d x

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

Solution steps

\int 6 x (x^{2} + 1)^{\frac{1}{2}} d x

Take the constant out:

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

= 6 \cdot \int x (x^{2} + 1)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{2} d u

= 6 \cdot \int \frac{u ^{\frac{1}{2}}}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int (3 x - sin (x)) d x

\frac{\partial}{\partial y} (2 x^{3} y - 3)

\frac{5 X ^{4} - 3 X ^{3}}{X ^{5}}

\frac{d ^{4}}{d x ^{4}} (9 e^{x^{2}})

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What is the integral of 6x(x^2+1)^{1/2} ?
The integral of 6x(x^2+1)^{1/2} is 2(x^2+1)^{3/2}+C