What is the integral of 5xln(sqrt(1+x)) ?

The integral of 5xln(sqrt(1+x)) is 5/8 (2x^2ln(x+1)-x^2+2x-2ln(x+1)+3)+C

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integral of 5xln(sqrt(1+x))

\int 5 x ln (1 + x) d x

\frac{5}{8} (2 x^{2} ln (x + 1) - x^{2} + 2 x - 2 ln (x + 1) + 3) + C

Solution steps

\int 5 x ln (1 + x) d x

Take the constant out:

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

= 5 \cdot \int x ln (1 + x) d x

Simplify

ln (1 + x) : \frac{1}{2} ln (x + 1)

= 5 \cdot \int \frac{1}{2} x ln (1 + x) d x

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int x ln (1 + x) d x

Apply u-substitution

: \int (u - 1) ln (u) d u

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

Apply Integration By Parts

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

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

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= 5 \cdot \frac{1}{2} (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

Substitute back

u = 1 + x

= 5 \cdot \frac{1}{2} (ln (1 + x) (\frac{( 1 + x ) ^{2}}{2} - (1 + x)) - (\frac{( 1 + x ) ^{2}}{4} - (1 + x)))

Simplify

5 \cdot \frac{1}{2} (ln (1 + x) (\frac{( 1 + x ) ^{2}}{2} - (1 + x)) - (\frac{( 1 + x ) ^{2}}{4} - (1 + x))) : \frac{5}{8} (2 x^{2} ln (x + 1) - x^{2} + 2 x - 2 ln (x + 1) + 3)

= \frac{5}{8} (2 x^{2} ln (x + 1) - x^{2} + 2 x - 2 ln (x + 1) + 3)

Add a constant to the solution

= \frac{5}{8} (2 x^{2} ln (x + 1) - x^{2} + 2 x - 2 ln (x + 1) + 3) + C

x^{2}, x

\frac{d}{d x} (8 sin (8 x))

\frac{d}{d x} (x^{\frac{1}{1}})

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

x \to + 0 + lim (\frac{x ^{2}}{4} - \frac{2}{x})

What is the integral of 5xln(sqrt(1+x)) ?
The integral of 5xln(sqrt(1+x)) is 5/8 (2x^2ln(x+1)-x^2+2x-2ln(x+1)+3)+C