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

The integral of ((x^2+4))/((x+3)) is ((x+3)^2)/2-6(x+3)+13ln|x+3|+C

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

integral

\frac{( x ^{2} + 4 )}{( x + 3 )}

\frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 13 ln ∣ x + 3 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} - 6 u + 13}{u} d u

= \int \frac{u ^{2} - 6 u + 13}{u} d u

Expand

\frac{u ^{2} - 6 u + 13}{u} : u - 6 + \frac{13}{u}

= \int u - 6 + \frac{13}{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

= \int u d u - \int 6 d u + \int \frac{13}{u} d u

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

\int 6 d u = 6 u

\int \frac{13}{u} d u = 13 ln ∣ u ∣

= \frac{u ^{2}}{2} - 6 u + 13 ln ∣ u ∣

Substitute back

u = x + 3

= \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 13 ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 13 ln ∣ x + 3 ∣ + C

f (x) = \frac{3 x + 3}{( 3 - 8 x + x ^{2} ) ^{\frac{1}{2}}}

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

x^{2} x \cdot 1

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

sin^{3} (x) cos^{2} (x)

What is the integral of ((x^2+4))/((x+3)) ?
The integral of ((x^2+4))/((x+3)) is ((x+3)^2)/2-6(x+3)+13ln|x+3|+C