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

The integral of x^3(a^2+x^2)^{1/2} is 1/15 (3x^2-2a^2)(x^2+a^2)^{3/2}+C

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integral of x^3(a^2+x^2)^{1/2}

\int x^{3} (a^{2} + x^{2})^{\frac{1}{2}} d x

\frac{1}{15} (3 x^{2} - 2 a^{2}) (x^{2} + a^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} (a^{2} + x^{2})^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{( u - a ^{2} ) u ^{\frac{1}{2}}}{2} d u

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

Take the constant out:

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

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

Expand

(u - a^{2}) u^{\frac{1}{2}} : u^{\frac{3}{2}} - a^{2} u^{\frac{1}{2}}

= \frac{1}{2} \cdot \int u^{\frac{3}{2}} - a^{2} u^{\frac{1}{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int u^{\frac{3}{2}} d u - \int a^{2} u^{\frac{1}{2}} d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int a^{2} u^{\frac{1}{2}} d u = a^{2} \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - a^{2} \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = a^{2} + x^{2}

= \frac{1}{2} (\frac{2}{5} (a^{2} + x^{2})^{\frac{5}{2}} - a^{2} \frac{2}{3} (a^{2} + x^{2})^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{2}{5} (a^{2} + x^{2})^{\frac{5}{2}} - a^{2} \frac{2}{3} (a^{2} + x^{2})^{\frac{3}{2}}) : \frac{1}{15} (3 x^{2} - 2 a^{2}) (x^{2} + a^{2})^{\frac{3}{2}}

= \frac{1}{15} (3 x^{2} - 2 a^{2}) (x^{2} + a^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 x^{2} - 2 a^{2}) (x^{2} + a^{2})^{\frac{3}{2}} + C

y = 6^{x}

y = 7 x, (7, 7)

\int \frac{x}{3 x ^{2} + 4} d x

f (x) = \frac{x}{2 x - 4}, at x = 5

n = 1 \sum \infty \frac{2}{1 + n ^{2}}

What is the integral of x^3(a^2+x^2)^{1/2} ?
The integral of x^3(a^2+x^2)^{1/2} is 1/15 (3x^2-2a^2)(x^2+a^2)^{3/2}+C