What is the integral of 2ylog_{10}(y)+1 ?

The integral of 2ylog_{10}(y)+1 is y^2log_{10}(y)-(y^2)/(2ln(10))+y+C

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integral of 2ylog_{10}(y)+1

integral

2 y lo g_{10} (y) + 1

y^{2} lo g_{10} (y) - \frac{y ^{2}}{2 ln ( 10 )} + y + C

Solution steps

\int 2 y lo g_{10} (y) + 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 y lo g_{10} (y) d y + \int 1 d y

\int 2 y lo g_{10} (y) d y = y^{2} lo g_{10} (y) - \frac{y ^{2}}{2 ln ( 10 )}

\int 1 d y = y

= y^{2} lo g_{10} (y) - \frac{y ^{2}}{2 ln ( 10 )} + y

Add a constant to the solution

= y^{2} lo g_{10} (y) - \frac{y ^{2}}{2 ln ( 10 )} + y + C

6 x \cdot e^{- 2} x

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

\frac{e ^{4} ln ( x ) - e ^{5} ln ( x )}{x ^{5}}

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

\frac{( cos ^{3} ( x ) )}{sin ^{4} ( x )}

What is the integral of 2ylog_{10}(y)+1 ?
The integral of 2ylog_{10}(y)+1 is y^2log_{10}(y)-(y^2)/(2ln(10))+y+C