What is the integral of ((x+8))/(sqrt(x+7)) ?

The integral of ((x+8))/(sqrt(x+7)) is 2(1/3 (x+7)^{3/2}+sqrt(x+7))+C

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integral of ((x+8))/(sqrt(x+7))

integral

\frac{( x + 8 )}{x + 7}

2 (\frac{1}{3} (x + 7)^{\frac{3}{2}} + x + 7) + C

Solution steps

\int \frac{x + 8}{x + 7} d x

Apply u-substitution

: \int 2 (u^{2} + 1) d u

= \int 2 (u^{2} + 1) d u

Take the constant out:

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

= 2 \cdot \int u^{2} + 1 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

= 2 (\int u^{2} d u + \int 1 d u)

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

\int 1 d u = u

= 2 (\frac{u ^{3}}{3} + u)

Substitute back

u = x + 7

= 2 (\frac{( x + 7 ) ^{3}}{3} + x + 7)

Simplify

2 (\frac{( x + 7 ) ^{3}}{3} + x + 7) : 2 (\frac{1}{3} (x + 7)^{\frac{3}{2}} + x + 7)

= 2 (\frac{1}{3} (x + 7)^{\frac{3}{2}} + x + 7)

Add a constant to the solution

= 2 (\frac{1}{3} (x + 7)^{\frac{3}{2}} + x + 7) + C

\frac{1}{x ^{2} + y ^{2}}

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

\frac{1}{1 + x ^{4}}

\frac{1}{( 1 - 2 x )}

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

What is the integral of ((x+8))/(sqrt(x+7)) ?
The integral of ((x+8))/(sqrt(x+7)) is 2(1/3 (x+7)^{3/2}+sqrt(x+7))+C