What is the integral of xln(4+x^4) ?

The integral of xln(4+x^4) is 1/2 (x^2ln(4+x^4)-4(-arctan((x^2)/2)+(x^2)/2))+C

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integral of xln(4+x^4)

\int x ln (4 + x^{4}) d x

\frac{1}{2} (x^{2} ln (4 + x^{4}) - 4 (- arctan (\frac{x ^{2}}{2}) + \frac{x ^{2}}{2})) + C

Solution steps

\int x ln (4 + x^{4}) d x

Apply u-substitution

: \int \frac{ln ( 4 + u ^{2} )}{2} d u

= \int \frac{ln ( 4 + u ^{2} )}{2} 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 ln (4 + u^{2}) d u

Apply Integration By Parts

: u ln (4 + u^{2}) - \int \frac{2 u ^{2}}{4 + u ^{2}} d u

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

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

= \frac{1}{2} (u ln (4 + u^{2}) - 4 (- arctan (\frac{u}{2}) + \frac{u}{2}))

Substitute back

u = x^{2}

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

(x^{2})^{2} = x^{4}

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

Add a constant to the solution

= \frac{1}{2} (x^{2} ln (4 + x^{4}) - 4 (- arctan (\frac{x ^{2}}{2}) + \frac{x ^{2}}{2})) + C

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What is the integral of xln(4+x^4) ?
The integral of xln(4+x^4) is 1/2 (x^2ln(4+x^4)-4(-arctan((x^2)/2)+(x^2)/2))+C