What is the integral of (x^3)/(sqrt(4x^2+25)) ?

The integral of (x^3)/(sqrt(4x^2+25)) is 1/24 (2x^2-25)sqrt(4x^2+25)+C

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integral of (x^3)/(sqrt(4x^2+25))

\int \frac{x ^{3}}{4 x ^{2} + 25} d x

\frac{1}{24} (2 x^{2} - 25) 4 x^{2} + 25 + C

Solution steps

\int \frac{x ^{3}}{4 x ^{2} + 25} d x

Apply u-substitution

: \int \frac{u ^{2} - 25}{16} d u

= \int \frac{u ^{2} - 25}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u^{2} - 25 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}{16} (\int u^{2} d u - \int 25 d u)

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

\int 25 d u = 25 u

= \frac{1}{16} (\frac{u ^{3}}{3} - 25 u)

Substitute back

u = 4 x^{2} + 25

= \frac{1}{16} (\frac{( 4 x ^{2} + 25 ) ^{3}}{3} - 25 4 x^{2} + 25)

Simplify

\frac{1}{16} (\frac{( 4 x ^{2} + 25 ) ^{3}}{3} - 25 4 x^{2} + 25) : \frac{1}{24} (2 x^{2} - 25) 4 x^{2} + 25

= \frac{1}{24} (2 x^{2} - 25) 4 x^{2} + 25

Add a constant to the solution

= \frac{1}{24} (2 x^{2} - 25) 4 x^{2} + 25 + C

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cos (x^{2})

\int \frac{sin ( \frac{1}{x ^{5}} )}{x ^{6}} d x

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What is the integral of (x^3)/(sqrt(4x^2+25)) ?
The integral of (x^3)/(sqrt(4x^2+25)) is 1/24 (2x^2-25)sqrt(4x^2+25)+C