Please answer 4 and 6
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I'll give the answer in brief.
For 4, put √{3x+4}=t and then it'll become 2(t²-8t+16)dt/9(t²+5), which can be integrated easily.
For 6, put √x=t and then it'll become 2dt/(t⁴+1), to solve this, write 2=(t²+1)-(t²-1) and then divide it into two integrals, i.e., t²+1/t⁴+1 and t²-1/t⁴+1 which we can solve by taking t² common from both denominator and numerator and then substitute t+1/t or t-1/t appropriately.
For 4, put √{3x+4}=t and then it'll become 2(t²-8t+16)dt/9(t²+5), which can be integrated easily.
For 6, put √x=t and then it'll become 2dt/(t⁴+1), to solve this, write 2=(t²+1)-(t²-1) and then divide it into two integrals, i.e., t²+1/t⁴+1 and t²-1/t⁴+1 which we can solve by taking t² common from both denominator and numerator and then substitute t+1/t or t-1/t appropriately.
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