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

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

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integral of 6xsqrt(1+x^2)

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

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

Solution steps

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

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

Simplify

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

y^{^{''}} + 20 y^{^{'}} + 500 y = 0

x d y + y d x = 0

\int π t - t^{2} cos (n t) d t

\frac{\frac{2}{3}}{s ^{2} + 2 s + 17}

\frac{25 - x ^{2}}{32 x}

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