What is the integral of 6xsqrt(x^2+1) ?

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

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integral of 6xsqrt(x^2+1)

\int 6 x x^{2} + 1 d x

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

Solution steps

\int 6 x x^{2} + 1 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 d x

Apply u-substitution

: \int u^{2} d u

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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