What is the integral of x^aln(bx) ?

The integral of x^aln(bx) is 1/(a+1)x^{a+1}ln(bx)-(x^{a+1})/((a+1)^2)+C

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integral of x^aln(bx)

\int x^{a} ln (b x) d x

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

Solution steps

\int x^{a} ln (b x) d x

Apply Integration By Parts

: \frac{1}{a + 1} x^{a + 1} ln (b x) - \int \frac{x ^{a}}{a + 1} d x

= \frac{1}{a + 1} x^{a + 1} ln (b x) - \int \frac{x ^{a}}{a + 1} d x

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

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

Add a constant to the solution

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

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What is the integral of x^aln(bx) ?
The integral of x^aln(bx) is 1/(a+1)x^{a+1}ln(bx)-(x^{a+1})/((a+1)^2)+C