What is the integral of 5xsqrt(2x^2+3) ?

The integral of 5xsqrt(2x^2+3) is 5/6 (2x^2+3)^{3/2}+C

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integral of 5xsqrt(2x^2+3)

\int 5 x 2 x^{2} + 3 d x

\frac{5}{6} (2 x^{2} + 3)^{\frac{3}{2}} + C

Solution steps

\int 5 x 2 x^{2} + 3 d x

Take the constant out:

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

= 5 \cdot \int x 2 x^{2} + 3 d x

Apply u-substitution

: \int \frac{u}{4} d u

= 5 \cdot \int \frac{u}{4} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= 5 \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 x^{2} + 3

= 5 \cdot \frac{1}{4} \cdot \frac{2}{3} (2 x^{2} + 3)^{\frac{3}{2}}

Simplify

5 \cdot \frac{1}{4} \cdot \frac{2}{3} (2 x^{2} + 3)^{\frac{3}{2}} : \frac{5}{6} (2 x^{2} + 3)^{\frac{3}{2}}

= \frac{5}{6} (2 x^{2} + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{5}{6} (2 x^{2} + 3)^{\frac{3}{2}} + C

y = \frac{2}{x}, y = 8 x, y = \frac{1}{8 x}

x \frac{d y}{d x} - y = x^{2} + y^{2}

f (x) = 2 x 2 x^{2} + 4

What is the integral of 5xsqrt(2x^2+3) ?
The integral of 5xsqrt(2x^2+3) is 5/6 (2x^2+3)^{3/2}+C