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

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

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

integral

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

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

Solution steps

\int 2 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

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

Apply u-substitution

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

Take the constant out:

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

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

Multiply by the conjugate of

u + 1 : \frac{u ^{2} - 1}{u - 1}

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

Apply u-substitution

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

Apply u-substitution

= 2 \cdot \frac{1}{2} \cdot \int w^{\frac{1}{2}} d w

Apply the Power Rule

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

in t e g r a l \frac{1}{- x}

in t e g r a l \frac{( y + 1 )}{( y - 1 )}

in t e g r a l (- 1)^{x}

in t e g r a l \frac{1}{( x ^{2} )}

in t e g r a l sin (2 x + 3)

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