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

The integral of (x^3)/(x^2+5) is 1/2 (x^2+5-5ln|x^2+5|)+C

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

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

\frac{1}{2} (x^{2} + 5 - 5 ln x^{2} + 5) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 5}{2 u} d u

= \int \frac{u - 5}{2 u} 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 \frac{u - 5}{u} d u

Expand

\frac{u - 5}{u} : 1 - \frac{5}{u}

= \frac{1}{2} \cdot \int 1 - \frac{5}{u} 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 1 d u - \int \frac{5}{u} d u)

\int 1 d u = u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= \frac{1}{2} (u - 5 ln ∣ u ∣)

Substitute back

u = x^{2} + 5

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

Add a constant to the solution

= \frac{1}{2} (x^{2} + 5 - 5 ln x^{2} + 5) + C

\frac{d}{d x} (\frac{d y}{d x}) = 4 x

f (x) = - 3 e^{- 2 x} + 3

\int \frac{1}{( x ^{2} + 2 ) ^{\frac{5}{2}}} d x

\int \frac{u - 1}{u ^{2}} d u

\int_{0}^{\infty} x e^{- x^{2}} d x

What is the integral of (x^3)/(x^2+5) ?
The integral of (x^3)/(x^2+5) is 1/2 (x^2+5-5ln|x^2+5|)+C