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The integral is the following:

$$\int_{0}^{1} t \ln(t^2+1) dt$$

I'm not sure whether to approach this by integration by parts or u-substitution, I've tried both but am getting stuck somewhere. Any hints to get me started would be greatly appreciated.

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$$\int_{0}^{1} t \ln(t^2+1) dt$$

HINT:

substitiute $z=t^2+1$ and $dz=2tdt$

$$=\frac 1 2\int \ln(z) dz$$

Now integrate by parts, $f=\ln(z),\quad df=\frac{1}{z}dz, \quad g=z, \quad dg=dz$

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Hint

$$ \int t \ln \left(t^2+1 \right)dt = \frac 12 \int \ln \left( t^2+1\right)\ d \left(t^2+1 \right) $$

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HINT:

Integrate by parts straightaway,

$$\int t\ln(1+t^2)\ dt=\ln(1+t^2)\int t\ dt-\int\left(\dfrac{d\ \ln(1+t^2)}{dt}\int t\ dt\right)dt$$

$$=\dfrac{t^2\ln(1+t^2)}2-\int\dfrac{t^3}{1+t^2}\ dt$$

Now, $$\int\dfrac{t^3}{1+t^2}\ dt=\int\dfrac{t(1+t^2)-t}{1+t^2}\ dt=\int t\ dt-\int\dfrac{t\ dt}{1+t^2}$$

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