What is the integral of 2xsqrt(x^2+1) ?

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

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integral of 2xsqrt(x^2+1)

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= 2 \cdot \int u^{2} d u

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 1

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

Simplify

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

2 y^{^{''}} - 54 y = 0, y (1) = 1, y (2) = 2

x \to 0 lim (x (ln (x)))

\int_{0}^{4} x^{2} d x

\frac{d}{d y} (x + y)

\int x^{2} 2 - x d x

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