Find:∫x4log xdxORFind: ∫2xx2+13dx

Let’s assume that I=∫x4log xdxBy using integration by parts,I=log x∫x4dx−∫ddxlog x∫x4dxdxWe know that ∫xndx=xn+1n+1 and ddxlog x=1x∴I=log⁡x×x55−∫1x×x55dx∴I=x55log⁡x−15∫x4dx∴I=x55log⁡x−15×x55∴I=x55log⁡x−x525+CTherefore, the value of ∫x4log xdx=x55log x−x525+CORFor given problem assume u=x2+1⇒du=2xdx⇒dx=12xdu∴∫2xx2+13dx=∫1u3du=∫1u13du=∫u−13du=u-13+1−13+1+C=u2323+C=32u23+CNow, substitute μ=x2+1∴∫2xx2+13dx=32x2+123+CWhich is the required answer.

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Find: $\int x^{4} \log xdx$
OR
Find: $\int \frac{2 x}{\sqrt[3]{x^{2} + 1}} dx$

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Detailed Solution

Let’s assume that $I = \int x^{4} \log xdx$
By using integration by parts,
$I = \log x \int x^{4} dx - \int (\frac{d}{dx} \log x \int x^{4} dx) dx$
We know that $\int x^{n} dx = \frac{x^{n + 1}}{n + 1} and \frac{d}{dx} \log x = \frac{1}{x}$
$\begin{array}{l} ∴ I = \log x \times \frac{x^{5}}{5} - \int \frac{1}{x} \times \frac{x^{5}}{5} dx \\ ∴ I = \frac{x^{5}}{5} \log x - \frac{1}{5} \int x^{4} dx \\ ∴ I = \frac{x^{5}}{5} \log x - \frac{1}{5} \times \frac{x^{5}}{5} \\ ∴ I = \frac{x^{5}}{5} \log x - \frac{x^{5}}{25} + C \end{array}$
Therefore, the value of $\int x^{4} \log xdx = \frac{x^{5}}{5} \log x - \frac{x^{5}}{25} + C$

For given problem assume $u = x^{2} + 1 \Rightarrow du = 2 xdx \Rightarrow dx = \frac{1}{2 x} du$
$\begin{array}{l} ∴ \int \frac{2 x}{\sqrt[3]{x^{2} + 1}} dx = \int \frac{1}{\sqrt[3]{u}} du \\ = \int \frac{1}{u^{\frac{1}{3}}} du \\ = \int u^{- \frac{1}{3}} du \\ = \frac{u^{\frac{- 1}{3} + 1}}{- \frac{1}{3} + 1} + C \\ = \frac{u^{\frac{2}{3}}}{\frac{2}{3}} + C \\ = \frac{3}{2} u^{\frac{2}{3}} + C \end{array}$
Now, substitute $μ = x^{2} + 1$
$∴ \int \frac{2 x}{\sqrt[3]{x^{2} + 1}} dx = \frac{3}{2} {(x^{2} + 1)}^{\frac{2}{3}} + C$
Which is the required answer.