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Question Number 42054 by SammyKT last updated on 17/Aug/18

Commented by math khazana by abdo last updated on 17/Aug/18

$$whatsC?Cisnotgiven!$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18

$$=\frac{1}{2}\int \frac{2z^2+z+2-z}{z(2z+1)}dz$$$$=\frac{1}{2}\int dz+\frac{1}{2}\int \frac{2-z}{z(2z+1)}dz$$$$=\frac{1}{2}\int dz+\frac{1}{2}\int \frac{z-(2z+1)+3}{z(2z+1)}dz$$$$=\frac{1}{2}\int dz+\frac{1}{2}\int \frac{dz}{2z+1}-\frac{1}{2}\int \frac{dz}{z}+\frac{3}{2}\int \frac{2z+1-2z}{z(2z+1)}dz$$$$=\frac{1}{2}\int dz+\frac{1}{4}\int \frac{dz}{z+\frac{1}{2}}-\frac{1}{2}\int \frac{dz}{z}+\frac{3}{2}\int \frac{dz}{z}-3\int \frac{dz}{2z+1}$$$$=\frac{1}{2}\int dz+\frac{1}{4}\int \frac{dz}{z+\frac{1}{2}}+\int \frac{dz}{z}-\frac{3}{2}\int \frac{dz}{z+\frac{1}{2}}$$$$=\frac{1}{2}z+\frac{1}{4}ln(z+\frac{1}{2})+lnz-\frac{3}{2}ln(z+\frac{1}{2})+c$$$$$$$$$$

Commented by SammyKT last updated on 17/Aug/18

$$Thankyousir...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18

$$itsok...$$

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