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

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

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

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

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

Solution steps

\int 3 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

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

Apply u-substitution

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

= 3 \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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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

csc (x) cot (x)

y = (sin (3 x))^{l n (x)}

\frac{\partial}{\partial y} (ln (y^{x}))

t \to - 5 lim (t^{2} - 5)

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