Question

${ \boxed{\bold \red{\int( {x}^{3} + 1) {}^{2} dx}}}$




❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​

Answer 1

Answer:

(2x³ + 2) 3x²

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Answer 2

$\bold \red{\int{\left(x^{3} + 1\right)^{2} d x}}$

${\color{red}{\int{\left(x^{3} + 1\right)^{2} d x}} = \color{red}{\int{\left(x^{6} + 2 x^{3} + 1\right)d x}}}$

${\color{red}{\int{\left(x^{6} + 2 x^{3} + 1\right)d x}} = \color{red}{\left(\int{1 d x} + \int{2 x^{3} d x} + \int{x^{6} d x}\right)}}$

${ \red{\int{2 x^{3} d x} + \int{x^{6} d x} + \color{red}{\int{1 d x}} = \int{2 x^{3} d x} + \int{x^{6} d x} + \color{red}{x}}}$

${x + \int{2 x^{3} d x} + \color{red}{\int{x^{6} d x}}=x + \int{2 x^{3} d x} + \color{red}{\frac{x^{1 + 6}}{1 + 6}}=x + \int{2 x^{3} d x} + \color{red}{\left(\frac{x^{7}}{7}\right)}}$

${\frac{x^{7}}{7} + x + \color{red}{\int{2 x^{3} d x}} = \frac{x^{7}}{7} + x + \color{red}{\left(2 \int{x^{3} d x}\right)}}$

${= \frac{x^{7}}{7} + x + 2 \color{red}{\frac{x^{1 + 3}}{1 + 3}}=\frac{x^{7}}{7} + x + 2 \color{red}{\left(\frac{x^{4}}{4}\right)}}$

Therefore,

${ \boxed{ \bold \red{\int{\left(x^{3} + 1\right)^{2} d x} = \frac{x^{7}}{7} + \frac{x^{4}}{2} + x+C}}}$