Integrate the function: \(\frac {3x}{1+2x^4}\)
Solution and Explanation
\(Let \ \sqrt 2x^2 = t\)
\(∴ 2\sqrt2x\ dx = dt\)
\(⇒ ∫\frac {3x}{1+2x^4}\ dx = \frac {3}{2\sqrt2} ∫\frac {dt}{1+t^2}\)
\(= \frac {3}{2√2}[tan^{-1} t]+C\)
\(= \frac {3}{2√2}[tan^{-1} (\sqrt 2x^2)]+C\)
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Questions Asked in CBSE CLASS XII exam
- Two coils ‘1’ and ‘2’ are placed close to each other as shown in the figure. Find the direction of induced current in coil ‘1’ in each of the following situations, justifying your answers:
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Bittu and Chintu were partners in a firm sharing profit and losses in the ratio of 4 : 3. Their Balance Sheet as at 31st March, 2024 was as follows:
On 1st April, 2024, Diya was admitted in the firm for \( \frac{1}{7} \)th share in the profits on the following terms:- (i) New profit sharing ratio between Bittu, Chintu and Diya was 3 : 3 : 1 .
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- (iii) Creditors were paid ₹ 4,20,000 in full settlement.
- (iv) Diya brought proportionate capital and ₹ 5,60,000 as her share of goodwill premium by cheque.
Prepare Revaluation Account and Partners' Capital Accounts.
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- Pass the necessary journal entries for the following transactions on the dissolution of a partnership firm of Vibha and Ajit after various assets (other than cash) and external liabilities have been transferred to Realisation Account:}
Creditors worth ₹ 46,000 accepted ₹ 9,000 cash and furniture of ₹ 32,000 in full settlement of their claim.
The firm had stock of ₹ 20,000. Ajit took over 40% of the stock at a discount of 10% while the remaining stock was sold for ₹ 18,000.
Vibha was appointed to look after dissolution work for which she was allowed a remuneration of ₹ 16,000. Vibha agreed to bear the dissolution expenses. Actual dissolution expenses ₹ 15,000 were paid by Vibha.
Ajit’s loan of ₹ 45,000 was settled at ₹ 42,000.
A machine which was not recorded in the books was taken over by Vibha at ₹ 23,000, whereas its expected value was ₹ 28,000.
The firm had a debit balance of ₹ 20,000 in the Profit and Loss Account on the date of dissolution.- CBSE CLASS XII - 2025
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- A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
